UNIVERSITY  OF  CALIFORNIA 
AT   LOS  ANGELES 


GIFT  OF 

Mrs.   Arthur   C.V.'oodv.^rc 


THE  ELECTEIC  CIRCUIT 


WORKS  BY  THE  SAME  AUTHOR 


Published  by  McGRAW-HILL  BOOK  COMPANY 

The  Electric  Circuit 

Svo,  xvi+229  pages.     Cloth $2.00 

(Sectmtt  Etlttwn,  entirely  rewritten  and  enlarged) 

The  Magnetic  Circuit 

Svo,  xviii+283  pages.     Cloth $2.00 


Published  by  JOHN  WILEY  &  SONS 

Experimental  Electrical  Engineering 

Vol.  I.    8vo,  xix+469  pages,  328  figures. 

Cloth net,         $3.50 

Vol.  II.   Svo,  xiv+333  pages,  209  figures. 

Cloth net,        $2.50 

Engineering  Applications  of  Higher  Mathe- 
matics 
Part  I.    MACHINE  DESIGN.     Small  Svo, 

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Published  by  FERDINAND  ENKE,  STUTTGART 
Ueber  Mehrphasige  Stromsysteme  bei 
Ungleichmassiger  Belastung.    Paper Mk.  2.40 


THE 

ELECTRIC    CIRCUIT 


BY 

V.   KARAPETOFF 


SECOND  EDITION 

REWRITTEN,  ENLARGED  AND  ENTIRELY  RESET 
SECOND  IMPRESSION 


McGRAW-HILL    BOOK    COMPANY 

239  WEST  39TH  STREET,  NEW  YORK 

6  BOUVERIE  STREET,  LONDON,  E.G. 

1912 


COPYRIGHT,  1912,  BY  THE 

McGRAW-HILL  BOOK  COMPANY 


COPYRIGHT,  19 10,  BY  V.  KAKAPKTOFF 


Stanhope  jprcss 

F.    H.GILSON   COMPANY 
BOSTON,  U.S.A. 


Engineering 
Lbruy 

-TK 

15-3 


PREFACE  TO   THE  FIRST  EDITION 


THIS  pamphlet,  together  with  the  companion  pamphlet  en- 
titled The  Magnetic  Circuit,  is  intended  to  give  a  student  in 
electrical  engineering  the  theoretical  elements  necessary  for  cal- 
culation of  the  performance  of  dynamo-electric  machinery  and 
of  transmission  lines.  The  advanced  student  must  be  taught  to 
treat  every  electric  machine  as  a  particular  combination  of  electric 
and  magnetic  circuits,  and  to  base  its  performance  upon  the 
fundamental  theoretical  relations  rather  than  upon  a  separate 
"  theory  "  established  for  each  kind  of  machinery,  as  is  often 
done. 

The  first  chapter  is  devoted  to  a  review  of  the  direct-current 
circuit,  the  next  four  chapters  treat  of  sine-wave  alternating- 
current  circuits,  and  the  last  two  chapters  give  the  fundamental 
properties  of  the  electrostatic  circuit.  All  the  important  results 
and  methods  are  illustrated  by  numerical  problems  of  which  there 
are  over  one  hundred  in  the  text.  The  pamphlet  is  not  intended 
for  a  beginner,  but  for  a  student  who  has  had  an  elementary  de- 
scriptive course  in  electrical  engineering  and  some  simple  labora* 
tory  experiments. 

The  treatment  is  made  as  far  as  possible  uniform,  so  that  the 
student  sees  analogous  relations  in  the  direct-current  circuit,  in 
the  alternating-current  circuit,  in  the  electrostatic  circuit  and 
finally  in  the  magnetic  circuit.  All  matter  of  purely  historical  or 
academic  interest,  not  bearing  directly  upon  the  theory  of  electric 
machinery,  has  been  left  out.  An  ambitious  student  will  find  a 
more  exhaustive  treatment  in  the  works  mentioned  at  the  end  of 
the  pamphlet. 

The  electrostatic  circuit  is  treated  in  accordance  with  the 
modern  conception  of  elastic  displacement  of  electricity  in  di- 
electrics. No  use  has  been  made  of  the  action  of  electric  charges 
at  a  distance,  or  of  the  electrostatic  system  of  units.  The  volt- 
ampere-ohm  system  of  units  is  used  for  electrostatic  calculations, 


488205 


yi  PREFACE  TO  THE  FIRST  EDITION 

in  accordance  with  Professor  Giorgi's  ideas  (see  a  paper  by  Pro- 
fessor Ascoli  in  Vol.  I  of  the  Transactions  of  the  International 
Electrical  Congress,  St.  Louis,  1904).  Those  familiar  with 
Oliver  Heaviside's  writings  will  notice  his  influence  upon  the 
author,  particularly  in  Arts.  22  and  23,*  where  an  attempt  is 
made  at  a  rational  electrostatic  nomenclature. 

Many  thanks  are  due  to  the  author's  friend  and  colleague, 
Mr.  John  F.  H.  Douglas,  instructor  in  electrical  engineering  in 
Sibley  College,  who  read  the  manuscript  and  the  proofs,  checked 
the  answers  to  the  problems  and  made  many  excellent  suggestions 
for  the  text. 

CORNELL  UNIVERSITY,  ITHACA,  N.  Y. 
August,  1910. 


PREFACE  TO  THE  SECOND  EDITION 


THE  first  edition  of  this  book  was  issued  as  a  pamphlet  of 
some  85  pages  which  the  author  used  for  two  years  in  his  classes 
to  supplement  some  other  texts.  In  its  present  edition,  the  book 
is  made  independent  of  these  texts,  so  that  its  size  had  to  be 
more  than  doubled.  The  book  has  been  practically  rewritten,  and 
completely  reset  in  type.  All  the  cuts  are  new.  The  topics  are 
treated  somewhat  more  in  detail,  and  a  large  number  of  practical 
problems  are  provided.  The  new  topics  added  are :  the  resistance 
of  conductors  of  variable  cross-section,  the  electrical  relations  in 
polyphase  systems,  performance  characteristics  of  the  trans- 
mission line,  transformer  and  induction  motor  and  the  permittance 
(electrostatic  capacity)  of  transmission  lines. 

In  the  treatment  of  alternating  currents  by  means  of  complex 
quantities,  particular  attention  is  paid  to  the  trigonometric  form 
E  (cos  8  +  j  sin  0)  of  the  expression  for  a  vector.  In  fact,  the 
transmission  line,  the  transformer,  and  the  induction  motor  to 
some  extent,  are  treated  in  this  trigonometric  form.  The  author 
trusts  that  the  reader  will  find  this  somewhat  novel  treatment 
more  convenient  in  numerical  applications  than  the  usual  form 
e+je'. 

*  Chapter  14  in  the  second  edition. 


PREFACE  TO  THE  SECOND  EDITION  vii 

Since  the  appearance  of  the  first  edition,  the  author  has  been 
encouraged  by  some  of  his  colleagues  in  his  treatment  of  the 
electrostatic  circuit  in  the  ampere-ohm  system  of  units,  a  treat- 
ment which  involves  the  use  of  permittances  in  farads  and 
elastances  in  darafs.  He  has  extended  this  treatment  to  the  calcu- 
lation of  capacity  of  cables  and  transmission  lines.  The  students 
grasp  this  mode  of  presentation  much  more  readily  than  the  old- 
fashioned  way,  based  upon  the  law  of  inverse  squares  and  elec- 
tric charges  acting  at  a  distance.  The  purpose  of  the  present 
treatment  is  to  impress  them  with  the  idea  of  a  continuous  action 
in  the  medium  itself  and  with  the  role  of  the  dielectric. 

Mr.  F.  R.  Keller  of  the  electrical  department  of  Columbia 
University  has  read  and  corrected  the  manuscript  and  the  proofs 
of  the  second  edition,  and  checked  the  answers  to  the  new  prob- 
lems. The  author  wishes  to  express  sincere  appreciation  of  his 
painstaking,  faithful  and  competent  work.  The  author  is  also 
indebted  to  Mr.  John  F.  H.  Douglas  for  critically  reading  the 
galley  proof  of  the  second  edition. 

CORNELL  UNIVERSITY,  ITHACA,  N.  Y. 
May,  1912. 


CONTENTS 


PREFACE v 

SUGGESTIONS  TO  TEACHERS xi 

CHAPTER  I.    FUNDAMENTAL  ELECTRICAL  RELATIONS  IN  DIRECT-CUR- 
RENT CIRCUITS 1 

The  volt,  the  ampere,  the  ohm,  and  the  mho.  Temperature  Co- 
efficient. Resistances  and  conductances  in  series  and  in  parallel. 
Electric  power. 

CHAPTER  II.     FUNDAMENTAL  ELECTRICAL  RELATIONS  IN  DIRECT-CUR- 
RENT CIRCUITS  (Continued) 13 

Resistivity  and  conductivity.  Current  density  and  voltage 
gradient.  Kirchhoff's  Laws. 

CHAPTER  III.     CONDUCTORS  OF  VARIABLE  CROSS-SECTION 22 

Current  density  and  voltage  gradient  at  a  point.     The  radial  flow 
of  current.     The  resistance  and  conductance  of  irregular  paths. 
The  law  of  current  refraction. 
CHAPTER    IV.     REPRESENTATION    OF    ALTERNATING    CURRENTS    AND 

VOLTAGES  BY  SINE-WAVES  AND  BY  VECTORS 31 

Sinusoidal  voltages  and  currents.  Representation  of  a  sine-wave 
by  a  vector.  Addition  and  subtraction  of  vectors.  Non-sinusoidal 
currents  and  voltages. 

CHAPTER  V.    POWER  IN  ALTERNATING-CURRENT  CIRCUITS 45 

Power  when  current  and  voltage  are  in  phase.  The  effective 
values  of  current  and  voltage.  Some  special  methods  for  calculating 
the  effective  value  of  an  irregular  curve.  Power  when  current  and 
voltage  are  out  of  phase. 

CHAPTER  VI.     INDUCTANCE,  REACTANCE,  AND  IMPEDANCE 60 

Inductance  as  electromagnetic  inertia.  Reactance.  Impedance. 
Influence  of  inductance  with  non-sinusoidal  voltage.  The  extra 
or  transient  current  in  opening  and  closing  a  circuit. 

CHAPTER  VII.     SUSCEPTANCE  AND  ADMITTANCE 75 

Concept  of  susceptance.  Concept  of  admittance.  Equivalent 
series  and  parallel  combinations.  Impedances  in  parallel  and  admit- 
tances in  series. 

CHAPTER  VIII.    THE  USE  OF  COMPLEX  QUANTITIES 82 

Addition  and  subtraction  of  projections  of  vectors.  Rotation  of 
vectors  by  ninety  degrees.  Impedance  and  admittance  expressed 
as  complex  quantities  or  operators. 

CHAPTER  IX.     THE  USE  OF  COMPLEX  QUANTITIES  (Continued) 91 

Power  and  phase  displacement  expressed  by  projections  of  vectors. 
Vectors  and  operators  in  polar  coordinates.  Vectors  and  operators 
expressed  as  exponential  functions. 


x  CONTENTS 

PAGE 

CHAPTER  X.    POLYPHASE  SYSTEMS 99 

Two-phase  system.  Three-phase  Y-connected  system.  Three- 
phase  delta-connected  system. 

CHAPTER  XI.     VOLTAGE  REGULATION  OF  THE  TRANSFORMER 108 

Imperfections  in  a  transformer  replaced  by  equivalent  resistances 
and  reactances.     The  vector  diagram  of  a  transformer.     Analytical 
determination  of  voltage  regulation.  —  Approximate  solution.    Ana- 
lytical determination  of  voltage  regulation.  —  Exact  solution. 
CHAPTER  XII.     PERFORMANCE  CHARACTERISTICS  OF  THE  INDUCTION 

MOTOR 122 

The  equivalent  electrical  diagram  of  an  induction  motor.     The 
analytical  determination  of  performance.  —  Approximate  solution. 
Starting  torque,  pull-out  torque,  and  maximum  output. 
CHAPTER  XIII.    PERFORMANCE  CHARACTERISTICS  OF  THE  INDUCTION 

MOTOR  (Continued) 133 

The  secondary  resistances  and  reactances  reduced  to  the  primary 
circuit.  The  circle  diagram.  The  analytical  determination  of  per- 
formance. —  Exact  solution. 

CHAPTER  XIV.    THE  DIELECTRIC  CIRCUIT 143 

The  electrostatic  field.  A  hydraulic  analogue  to  the  dielectric 
circuit.  The  permittance  and  elastance  of  dielectric  paths.  Per- 
mittivity and  elastivity  of  dielectrics.  Dielectric  flux  density  and 
electrostatic  stress  (voltage  gradient). 

CHAPTER  XV.    THE  DIELECTRIC  CIRCUIT  (Continued) 157 

Energy  in  the  electrostatic  field.     The  permittance  and  elastance 
of   irregular  paths.    The  law  of  flux  refraction.     The  dielectric 
strength  of  insulating  materials.     The  electrostatic  corona.     Dielec- 
tric hysteresis  and  conductance. 
CHAPTER    XVI.     ELASTANCE    AND    PERMITTANCE    OF    SINGLE-PHASE 

CABLES  AND  TRANSMISSION  LINES 171 

The  elastance  of  a  single-core  cable.  The  elastance  of  a  single- 
phasc  line.  The  influence  of  the  ground  upon  the  elastance  of  a 
single-phase  line.  The  equations  of  the  electrostatic  lines  of  force 
and  equipotential  surfaces  produced  by  a  single-phase  line.  The 
elastanre  between  two  large  parallel  circular  cylinders. 
CHAPTER  XVII.  EQUIVALENT  ELASTANCE  AND  CHARGING  CURRENT  IN 

THREE-PHASE  LINES 193 

Three-phase  line  with  symmetrical  spacing.     Three-phase  line 
with  unsymmetrical  spacing. 
CHAPTER    XVIII.    DIELECTRIC    REACTANCE    AND    SUSCEPTANCE    IN 

ALTERNATING-CURRENT  CIRCUITS 203 

Dielectric  reactance  and  susceptance.  Current  and  voltage 
resonance.  Voltage  regulation  of  a  transmission  line,  taking  its  dis- 
tributed permittance  into  account.  Approximate  formula;  for  the 
voltage  regulation  of  a  transmission  line,  considering  its  permittance 
concentrated  at  one  or  more  points. 

APPENDIX 215 

BIBLIOGRAPHY ' 


SUGGESTIONS   TO   TEACHERS 


(1)  THIS  book  is  intended  to  be  used  as  a  text  in  a  course 
which   comprises   lectures,    recitations,    computing   periods   and 
home  work.     Purely  descriptive  matter  has  been  omitted  or  only 
suggested,  in  order  to  allow  the  teacher  more  freedom  in  his 
lectures  and  to  permit  him  to  establish  his  own  point  of  view. 
Some  parts  of  the  book  are  more  suitable  for  recitations,  others  for 
reference  in  the  designing  room,  others  again  as  a  basis  for  dis- 
cussion in  the  lecture  room,  or  for  brief  theses. 

(2)  Different  parts  of  the  book  are  made  as  much  as  possible 
independent  of  one  another,  so  that  the  teacher  can  schedule 
them  as  it  suits  him  best.     Moreover,  most  chapters  are  written 
according  to  the  concentric  method,  so  that  it  is  not  necessary  to 
finish  one  chapter  before  starting  on  the  next.     One  can  cover  the 
subject  in  an  abridged  manner,  omitting  the  last  parts  of  some 
chapters. 

(3)  The  problems  given  at  the  end  of  nearly  every  article  are 
an  integral  part  of  the  book,  and  should  under  no  circumstances 
be  omitted.     There  is  no  royal  way  of  obtaining  a  clear  under- 
standing of  the  underlying  physical  principles,  and  of  acquiring 
an  assurance  in  their  practical  application,  except  by  the  solution 
of  numerical  examples. 

(4)  The  book  contains  comparatively  few  sketches,  in  order  to 
give  the  student  an  opportunity  to  illustrate  the  important  re- 
lations by  sketches  of  his  own.     Making  sketches,  diagrams  and 
drawings  of  electric  circuits  and  machines  to  scale  should  be  one 
of  the  important  features  of  the  course,  even  though  it  may  not 
be  popular  with  some  analytically  inclined  students.     Mechanical 
drawing  develops  precision  of  judgment,  and  gives  the  student  a 
knowledge  that  is  tangible  and  concrete. 

(5)  The  author  has  avoided  giving  definite  numerical  data, 
coefficients  and  standards,  except  in  problems,  where  they  are  in- 
dispensable and  where  no  general  significance  is  ascribed  to  such 


Xll 


SUGGESTIONS  TO  TEACHERS 


data.  His  reasons  are:  (a)  Numerical  coefficients  obscure  the 
general  exposition.  (6)  Sufficient  numerical  coefficients  and  de- 
sign data  will  be  found  in  good  electrical  hand-books  and  pocket- 
lxx)ks,  one  of  which  ought  to  be  used  in  conjunction  with  this 
text,  (c)  The  student  is  likely  to  ascribe  too  much  authority  to 
a  numerical  value  given  in  a  text-book,  while  in  reality  many 
coefficients  vary  within  wide  limits  according  to  the  conditions  of 
a  practical  problem,  and  with  the  progress  of  the  art.  (d)  Most 
numerical  coefficients  are  obtained  in  practice  by  assuming  that 
the  phenomenon  in  question  occurs  according  to  a  definite  law, 
and  by  substituting  the  available  experimental  data  into  the  cor- 
responding formula.  This  point  of  view  is  emphasized  through- 
out the  book,  and  gives  the  student  the  comforting  feeling  that  he 
will  be  able  to  obtain  the  necessary  numerical  constants  when 
confronted  by  a  definite  practical  situation. 

(6)  The  treatment  of  the  electrostatic  circuit  is  made  as 
much  as  possible  analogous  to  that  of  the  electrodynamic  cir- 
cuit. The  teacher  will  find  it  advisable  to  make  his  students 
perfectly  familiar  with  the  juse  of  Ohm's  law  for  ordinary  electric 
circuits  before  starting  on  the  electrostatic  circuit.  The  student 
should  solve  several  numerical  examples  involving  voltages  and 
voltage  gradients,  currents  and  current  densities,  resistances,  re- 
sistivities, conductances  and  conductivities.  He  will  then  find 
very  little  difficulty  in  mastering  the  electrostatic  circuit,  and 
from  these  two  the  transition  to  the  magnetic  circuit,  treated  in 
the  companion  book,  is  very  simple  indeed.  The  following  table 
shows  the  analogous  quantities  in  the  three  kinds  of  circuits. 


Electrodynamic 
i  Voltage  or  c.m.f. 
j  Voltage  gradient  (or 
electric  intensity) 

Electrostatic 
Voltage  or  e.m.f. 
Voltage    gradient    (or 
electric  intensity) 

Magnetic 
Magnetomotive  force 
M.m.f.  gradient  (or  mag- 
netic intensity) 

I  Electric  current 
i  Current  density 

Dielectric  flux 
Dielectric  flux  density 

Magnetic  flux 
Magnetic  flux  density 

t  Resistor 
<  Resistance 
'  Resistivity 

Elastor 
Elastance 
Elastivity 

Reluctor 
Reluctance 
Reluctivity 

i  Conductor 
<  Conductance 
'  Conductivity 

Permittor   (condenser) 
Permittance  (capacity) 
Permittivity      (dielec- 
tric constant) 

Permeator 
Permeance 
Permeability 

LIST  OF  PRINCIPAL  SYMBOLS. 

The  following  list  comprises  most  of  the  symbols  used  in  the  text.  Those 
not  occurring  here  are  explained  where  they  appear.  When,  also,  a  symbol 
has  a  use  different  from  that  stated  below,  the  correct  meaning  is  given  where 

the  symbol  occurs. 

Page  where  defined 
Symbol.  Meaning.  or  first  used. 

a         Radius  of  conductor  of  transmission  line 176 

a        Radius  of  core  of  cable 171 

A       Cross-section 13 

b         Inter  axial  distance  between  conductors  of  transmission  line 176 

6         Radius  of  inner  surface  of  cable  sheathing 171 

b        Susceptance 75 

C        Constant 54,  72 

C        Permittance  or  electrostatic  capacity 147 

D        Dielectric  flux  density 154 

Dmax  Rupturing  flux  density 165 

e         Electromotive  force 1 

e         Instantaneous  value  of  voltage 34 

e         Horizontal  component  of  vector  of  e.m.f 83 

e'        Vertical  component  of  vector  of  e.m.f 83 

ei        Local  source  of  e.m.f 3 

et        Terminal  voltage 3 

E        Effective  value  of  alternating  voltage 48 

E        Vector  of  the  voltage  E 83 

Em     Maximum  value  of  voltage 34 

/         Frequency  of  alternating  current  or  voltage 33 

F        Mechanical  force 60 

g         Conductance 2 

geq     Equivalent  conductance 8 

G        Voltage  gradient  or  electric  intensity 16 

Gmax  Rupturing  voltage  gradient 165 

h        Elevation  of  conductor  above  the  ground 181 

h        Head  of  fluid 10 

h        Instantaneous  value  of  harmonic 54 

i         Current 1 

i         Horizontal  component  of  vector  of  current 87 

i         Instantaneous  value  of  current 33 

i'        Vertical  component  of  vector  of  current 87 

7       Effective  value  of  alternating  current 48 

/       Vector  of  the  current  / 88 


xiv  LIST  OF  PRINCIPAL  SYMBOLS 

Page  where  defined 
Symbol.  Meaning.  or  first  used. 

IL  Primary  load  current 1 16 

Im  Maximum  value  of  current 34 

/„  Mesh  currents  in  squirrel-cage  rotor 133 

j  V^T 83.  85 

kb  Breadth  factor  of  winding 133 

A"  Relative  permittivity 151 

/  Length 13 

log  Common  logarithm 172 

L  Inductance - 60 

Ln  Natural  logarithm 171 

m  Mass 60 

m  Number  of  phases 133 

p  Number  of  poles 135 

P  Constant 73 

P  Input  per  phase  of  induction  motor 123 

P  Power 10 

Pate  Average  power 48 

q  Instantaneous  displacement  of  electricity 193 

q  Rate  of  discharge  of  a  fluid 10 

Q  Constant 73 

Q  Quantity  of  electricity 144 

Q  Quantity  of  heat '. 2 

r  Resistance 1 

rtq  Equivalent  resistance 7 

R  Resistance 8 

Ro  Resistance  at  0°  C 5 

Rt  Resistance  at  1°  C 5 

«  Slip  of  induction  motor 123 

<S  Area  of  curve 53 

S  Elastance 148 

t  Time 33 

T  Temperature 6 

T  Time  of  one  cycle  of  alternating  wave 33 

u  Variable  angle 33 

U  Current  density 15 

v  Velocity 60 

V  Volume ......."".'I.'!!!!!!'.!!  15» 

W  Energy \\\\  46 

W  Density  of  energy 158 

x  Reactance 53 

x  Variable  radius 171 

y  Admittance 76 

y  Ordinate  of  curve 50 

Y  Admittance  operator 89 

z  Impedance 67 

Z  Impedance  operator 88 


LIST  OF  PRINCIPAL  SYMBOLS  XV 

Page  where  defined 
Symbol.  Meaning.  or  first  used. 

a  Angle 43,  94 

a  A  ratio 189 

a  Temperature  coefficient 5 

7  Conductivity 14 

e  Base  of  natural  system  of  logarithms 72 

6  Difference  of  temperature 2 

6  Phase  angle 82 

01  Angle  of  incidence  of  current 28 

61  Angle  of  incidence  of  dielectric  flux 163 

02  Angle  of  refraction  of  current 28 

02  Angle  of  refraction  of  dielectric  flux 163 

K  Permittivity 151 

Ka  Permittivity  of  air 151 

p  Resistivity 13 

<r  Circle  coefficient  or  dispersion  factor  of  induction  motor 138 

a  Elastivity 152 

a a  Elastivity  of  air 152 

T  Time  constant  of  a  circuit 72 

<£  Phase  angle 34 

0  Magnetic  flux 114 

$m  Maximum  value  of  magnetic  flux 114 

$  Angle 121 

w  Angle 52 

O  Angle 52 


THE  BLBCTEIC  CIRCUIT 


CHAPTER  I 

FUNDAMENTAL  ELECTRICAL  RELATIONS  IN  DIRECT- 
CURRENT  CIRCUITS 

1.  The  Volt,  the  Ampere,  the  Ohm,  and  the  Mho.    The 

student  is  supposed  to  be  familiar  with  Ohm's  law,  both  theoret- 
ically and  from  his  laboratory  experience.  A  brief  synopsis  of 
the  law,  given  below,  is  intended  to  refresh  the  relations  in  his 
mind,  and  to  establish  a  point  of  view  which  permits  of  extending 
these  relations  to  alternating-current  circuits.  Moreover,  the 
law  is  presented  in  a  form  applicable  to  magnetic  and  dielectric 
circuits. 

When  the  current  in  a  conductor  is  steady  and  there  are  no 
local  electromotive  forces  within  the  conductor,  the  value  of  the 
current  is  proportional  to  the  voltage  between  the  terminals  of 
the  conductor.  This  is  an  experimental  fact,  called  Ohm's  law. 
The  word  "conductor"  is  used  here  in  the  sense  of  "the  part  of 
the  circuit  under  consideration. "  It  may  consist  of  two  or  more 
distinct  physical  conductors.  Considering  the  electromotive  force 
e  as  the  cause  of  the  current  i,  this  law  merely  states  that  the  effect 
is  proportional  to  the  cause,  or 

e  =  r-i, (1) 

where  the  coefficient  of  proportionality  r  is  called  the  resistance 
of  the  conductor.  When  the  current  is  expressed  in  amperes, 
and  the  electromotive  force  in  volts,  the  resistance  r  is  measured 
in  units  called  ohms. 

Ohm's  law  is  sometimes  written  in  the  form 

i  =  9-e, (2) 


2  THE  ELECTRIC  CIRCUIT  [ART.  1 

where  the  coefficient  of  proportionality 

g  =  l/r (3) 

is  called  the  conductance  of  the  conductor.  The  reason  for  this 
name  is  easy  to  see:  The  resistance  r  shows  how  difficult  it  is  to 
force  a  unit  current  through  a  given  conductor,  while  its  recip- 
rocal g  shows  how  easy  it  is  to  produce  the  same  current  in  the 
same  conductor.  Conductances  are  measured  in  units  called 
mhos,  one  mho  being  the  reciprocal  of  one  ohm.  Hence,  a  resist- 
ance of  one  ohm  represents  at  the  same  time  a  conductance  of 
one  mho;  a  resistance  of  two  ohms  has  a  conductance  of  one-half 
mho,  etc.  Increasing  the  resistance  of  a  winding  from  4  to  5  ohms 
reduces  its  conductance  from  0.25  to  0.20  mho. 

It  will  be  seen  below  that  in  some  problems  it  is  convenient  to 
use  conductances  instead  of  resistances.  Both  are  fundamental, 
and  there  is  no  reason  why  Ohm's  law  should  not  have  been  ex- 
pressed originally  by  eq.  (2)  instead  of  (1). 

With  our  present  meager  knowledge  of  the  true  nature  of 
electrical  phenomena,  it  is  well-nigh  impossible  to  give  a  clear 
physical  meaning  of  the  quantities  under  discussion  without 
resorting  to  analogies.  For  instance,  the  flow  of  current  through 
a  conductor  may  be  compared  to  the  flow  of  heat  through  a 
rod;  the  voltage  or  difference  of  electric  potential  is  analogous 
to  the  difference  of  temperature  6  at  the  ends  of  the  rod,  and  the 
electric  current  to  the  quantity  of  heat  Q  passing  through  a  cross- 
section  of  the  rod  in  unit  time  (the  rate  of  flow  of  heat).  The 
ratio  of  0  to  Q  is  sometimes  called  the  thermal  resistance  of 
the  rod.1 

Again,  the  phenomenon  of  the  flow  of  electricity  is  somewhat 
analogous  to  the  flow  of  water  through  pipes.  The  hydraulic 
head  may  be  likened  to  the  voltage,  and  the  rate  of  discharge  of 
water  to  the  current.  With  very  low  velocities,  in  capillary 
tubes,  the  discharge  is  proportional  to  the  head,  so  that  eqs.  (1) 
and  (2)  hold  true  for  the  flow  of  water. 

Whatever  the  reasons  which  have  led  originally  to  the  choice 
of  the  magnitudes  of  the  ampere,  the  ohm,  and  the  volt,  these 
units  may  be  considered  at  present,  for  all  practical  and  most 
theoretical  purposes,  as  arbitrary  units,  like  the  foot,  the  pound,  or 

1  It  even  has  been  proposed  to  measure  this  resistance  in  thermal  ohms  or 
thohm*. 


CHAP.  I]  DIRECT-CURRENT  CIRCUITS  3 

the  meter.  Their  values  have  been  established  by  an  interna- 
tional agreement,  whence  the  name,  international  electrical  units. 
These  units  are  represented  by  concrete  standards  with  minutely 
specified  dimensions  and  properties;  the  ohm  by  a  column  of 
mercury,  the  ampere  by  a  silver  voltameter,  and  the  volt  by  a 
standard  cell.  It  is  understood,  of  course,  that  only  two  out  of 
the  three  units  need  to  be  standardized,  the  third  being  determined 
either  as  their  product,  or  their  ratio.  It  has  been  decided  by 
international  agreement  to  consider  the  ampere  and  the  ohm  as 
fundamental  units,  the  volt  being  derived  from  them.  Hence 
the  present  system  of  practical  electrical  units  is  properly  called 
the  ampere-ohm  system.  This  fact  does  not  preclude,  of  course, 
the  use  of  standard  cells  as  secondary  standards. 

The  ampere,  the  volt,  and  the  ohm  are  connected  by  simple 
multipliers  (powers  of  10)  with  the  absolute  electromagnetic  units 
(the  C.G.S.  system  of  units).  It  is  conceded  at  present  by  some 
prominent  physicists  that  the  choice  of  the  units  was  not  quite 
fortunate,  according  to  our  present  understanding  of  the  electro- 
magnetic relations.  Since,  however,  it  is  too  late  to  change  these 
units,  it  is  better  to  consider  them  as  arbitrary,  and  not  con- 
nected in  any  way  with  the  magnitudes  of  the  centimeter,  the 
gram,  and  the  second. 

In  applying  Ohm's  law  to  practical  problems,  it  must  be 
clearly  remembered  that  e  represents  the  net  voltage  acting  be- 
tween the  ends  of  the  conductor  r.  This  is  important  when  the 
circuit  contains  sources  of  counter-electromotive  force,  such  as 
electric  batteries,  or  motors.  Let,  for  instance,  the  total  resistance 
of  a  circuit,  connected  across  the  terminals  of  a  generator,  be  12 
ohms,  and  let  the  terminal  voltage  of  the  generator  be  120  volts. 
Then  the  current  is  equal  to  10  amperes,  provided  that  there  are 
no  counter-electromotive  forces  in  the  circuit.  Let,  however,  the 
circuit  contain  a  storage  battery  of,  say,  24  volts,  connected  so 
as  to  be  charging,  that  is,  opposing  the  applied  voltage.  The 
current  in  the  circuit  is  now  only  (120  —  24)/12  =  8  amp.,  the 
value  120  —  24  =  96  being  the  net  voltage  in  the  external  circuit. 
Should  the  terminals  of  the  battery  be  reversed,  so  as  to  help  the 
generator  voltage,  the  current  would  increase  to  (120  +  24)/12 
=  12  amp. 

Thus,  when  there  is  an  external  or  local  source  of  electro- 
motive force,  say  ei,  within  a  conductor,  the  terminal  voltage  et 


4  THE  ELECTRIC  CIRCUIT  [ART.  1 

between  the  ends  of  the  conductor  is  added  algebraically  to  eh 
BO  that  we  have,  instead  of  eq.  (1), 

et  +  ei=i-r, (4) 

where  e\  is  considered  positive  when  in  the  same  direction  as  et. 
In  the  foregoing  numerical  example  the  counter-e.m.f.  is  therefore 
considered  negative. 

In  numerical  computations  it  is  sometimes  convenient  to  use 
multiples  and  submultiples  of  the  units  originally  agreed  upon, 
in  order  to  avoid  large  numbers  or  very  small  fractions.  This  is 
accomplished  by  adding  to  the  names  of  the  original  units  certain 
Greek  prefixes  for  the  multiples,  and  Latin  prefixes  for  the  sub- 
multiples.  These  prefixes  are  as  follows: 

deca. . .  .ten  deci one  tenth 

hecto. .  .one  hundred  centi one  hundredth 

kilo one  thousand  milli one  thousandth 

mega.  .  .one  million  micro one  millionth. 

For  instance,  instead  of  10,000  amperes  one  may  say  or  write 
10  kiloamperes;  instead  of  0.0003  volt  one  may  say  0.3  millivolt, 
or  300  microvolts,  etc.  Another  way  to  avoid  very  large  or  very 
small  numbers  is  to  use  10  to  the  proper  power  as  a  multiplier. 
For  instance,  one  may  speak  of  a  resistance  of  7  X  10~6  ohm,  of  a 
conductance  equal  to  5  X  107  mhos,  etc. 

Prob.  1.  In  order  to  determine  the  resistance  of  the  armature  of  an 
electrical  machine,  a  direct  current  is  sent  through  it  and  the  drop  of  volt- 
age is  measured  between  the  brushes.  The  following  are  the  readings: 

Volts 0.44         0.73          1.00          1.33         1.73 

Amperes 8.1          12.9          18.1          24.0          31.0 

What  is  the  most  probable  value  of  the  resistance?  Hint:  Take  an 
average  of  the  ratios,  or  better,  plot  the  volts  against  the  amperes  as 
abscissa?  and  draw  a  straight  line  through  the  origin. 

Ans.     0.0559  ohm. 

Prob.  2.  The  resistance  of  a  transmission  line  is  1.2  ohms.  What 
voltage  is  necessary  at  the  generating  end  in  order  to  produce  a  current 
of  75  amp.  (a)  when  the  line  is  short-circuited  at  the  receiving  end; 
(b)  when  a  pressure  of  500  volts  must  be  maintained  at  the  receiving 
e"d?  Ans.  90  volts;  590  volts. 

Prob.  3.  The  armature  resistance  of  a  250-volt  generator  is  0.025 
ohm.  At  what  current  will  the  voltage  drop  in  the  armature  be  equal 
to  4  per  cent  of  the  terminal  voltage?  Ans.  400  amp. 


CHAP.  I] 


DIRECT-CURRENT  CIRCUITS 


Prob.  4.  The  conductance  of  a  bath  of  molten  metal  is  5  kilomhos; 
what  voltage  is  required  to  send  a  current  of  7  X  104  amp.  through  it? 

Ans.     14  volts. 

Prob.  5.  The  coil  of  a  regulating  electromagnet  of  50  ohms  resistance 
is  connected  across  a  110- volt  line;  the  voltage  of  the  line  fluctuates  by 
±  10  per  cent.  In  order  to  make  the  regulating  mechanism  more  sensi- 
tive, that  is,  in  order  to  accentuate  the  fluctuations  of  current  in  it,  a 
counter-e.m.f.  storage  battery  of  negligible  resistance  is  connected  in 
series  with  the  coil.  What  must  be  the  voltage  of  the  battery  if  the  cur- 
rent in  the  coil  at  120  volts  must  be  twice  that  at  100  volts? 

Ans.    80  volts. 

2.  Temperature  Coefficient.  The  resistance  of  all  metals  and 
of  practically  all  alloys  increases  with  the  temperature,  according 
to  a  rather  complicated  law.  Within  the  usual  limits  of  tem- 
perature the  increase  in  resistance  is  nearly  proportional  to  the 
temperature  rise;  in  other  words,  the  relation  between  the  resist- 
ance of  a  conductor  and  its  temperature  is  represented  by  a  straight 


FIG.  1.    The  relation  between  the  resistance  and  the  temperature  of 
metals. 

line  MN  (Fig.  1).  Let  the  resistance  at  0°  C.  be  R0  ohms;  then 
the  resistance  at  some  temperature  t°  C.  is 

Rt  =  R0(l+a£), .     (5; 

where  a  is  called  the  temperature  coefficient  of  the  material.  For 
the  values  of  a  for  various  materials  see  an  electrical  handbook. 
For  the  most  important  material,  copper,  a  —  0.0042;  in  other 
words,  the  resistance  of  a  copper  conductor  increases  by  0.42  per 
cent  for  each  degree  centigrade,  considering  the  resistance  at 
0°  C.  as  100  per  cent. 

The  formula  given  below  is  sometimes  more  convenient  in 
computations  than  formula  (5).     Assume  that  the  same  straight- 


6  THE  ELECTRIC  CIRCUIT  [ART.  2 

line  law  (Fig.  1)  holds  for  low  temperatures,  and  let  temperatures 
be  measured  from  the  point  A  at  which  the  straight  line  crosses 
the  axis  of  abscissae.  Denoting  temperatures  from  this  point  by 
T,  we  have,  for  any  two  temperatures, 

Rl/R2  =  Tt/Tt (6) 

With  this  formula  it  is  not  necessary  to  refer  computations  to 
the  resistance  at  0°  C.  The  point  A  is  found  from  the  condition 
Rt  =  0,  from  which,  according  to  eq.  (5),  tA  =  —I/a.  Thus,  for 

any  temperature, 

T  =  t  +  l/a (7) 

For  copper,  T  =  t  +  238.1;  that  is,  point  A  lies  238.1°  C.  below 
the  freezing  point  of  water.  This  does  not  mean  that  the  resist- 
ance of  copper  actually  varies  according  to  this  law  at  such 
temperatures;  A  is  merely  a  fictitious  point  through  the  position 
of  which  it  is  convenient  to  express  the  equation  of  the  full-drawn 
part  of  the  straight  line  in  Fig.  1. 

Let,  for  instance,  the  resistance  of  the  winding  of  an  electric 
machine  be  0.437  ohm  at  the  room  temperature  of  22°  C.  After 
the  machine  has  been  run  for  several  hours  the  resistance  of  the 
same  winding  is  found  to  be  0.482  ohm,  with  the  room  temper- 
ature unchanged.  Let  it  be  required  to  calculate  the  final  tem- 
perature of  the  winding  from  the  increase  in  its  resistance.  We 
have  Ti  =  238.1  +  22  =  260.1,  and  according  to  eq.  (6)  the  un- 
known final  temperature  T2  =  260.1  X  (482/437)  =  286.9,  or 
tt  =  286.9  -  238.1  =  48.8°  C.  Two  other  practical  formulae  for 
temperature  rise  will  be  found  in  Appendix  E  of  the  Standard- 
ization Rules  of  the  American  Institute  of  Electrical  Engineers. 
These  rules  are  reprinted  in  most  electrical  handbooks  and  pocket- 
books.  See  also  a  convenient  method  given  in  problem  2  below. 

Prob.  1.  The  resistance  of  a  conductor  increases  by  31  per  cent  from 
23°  to  75°  C.  What  is  «  in  formula  (5)?  Ans.  0.00691. 

Prob.  2.  The  relation  between  the  resistance  and  the  temperature 
of  copper  conductors  is  easily  obtained  on  an  ordinary  slide  rule,  as  fol- 
lows: On  the  lower  movable  scale  mark  0°  C.  on  division  238;  10°  C.  on 
division  248,  and  so  on.  Set  a  known  resistance  on  the  lower  fixed  scale, 
and  bring  the  corresponding  temperature  opposite.  Then  the  resistance 
at  any  other  temperature  is  read  opposite  the  corresponding  division  on 
the  temperature  scale.  Give  an  explanation  of  this  method. 

Prob.  3.  Prove  the  formulae  for  #<+,  and  r  in  the  above-mentioned 
Standardization  Rules. 


CHAP.  I]  DIRECT-CURRENT  CIRCUITS  7 

3.  Resistances  and  Conductances  in  Series  and  in  Parallel. 

When  resistances  are  connected  in  series,  the  total  resistance  of 
the  circuit  is  increased.  This  can  be  more  easily  seen  by  resort- 
ing to  analogies.  For  instance,  if  the  length  of  a  pipe  carrying  a 
fluid  be  increased,  the  frictional  resistance  to  the  flow  becomes 
greater;  in  like  manner,  a  long  rod  offers  a  more  difficult  path  for 
the  passage  of  heat  than  a  short  one.  In  the  electric  circuit,  the 
equivalent  resistance  of  two  conductors  in  series  is  equal  to  the 
sum  of  their  individual  resistances,  as  is  shown  below.  This  follows 
from  the  experimental  fact  that  electricity  in  its  flow  behaves  like 
an  incompressible  fluid;  that  is,  the  same  quantity  of  it  must  pass 
in  a  given  time  interval  through  all  the  cross-sections  of  a  circuit. 
Let  two  conductors,  r\  and  r^,  be  connected  in  series  across  a 
source  of  voltage  e,  and  let  a  current  i  flow  through  them.  Part 
of  the  total  voltage  e  is  spent  in  overcoming  the  resistance  of  the 
first  conductor,  the  rest  in  overcoming  that  of  the  second  con- 
ductor. But,  according  to  Ohm's  law,  when  the  conditions  are 
steady,  the  voltage  across  the  first  conductor,  ei  =  i-n;  the 
voltage  across  the  second  is  ez  =  i  •  r%.  Adding  these  two  equa- 
tions gives  the  total  voltage 

e  =  ei  +  e2  =  i(ri  +  r2). 

An  equivalent  resistance,  req,  by  definition,  is  one  which,  with 
the  same  total  voltage  e,  allows  the  same  current  i  to  pass  through 
the  circuit,  as  the  combination  of  the  given  conductors.  Hence, 

e  =  i-  req. 
Comparing  the  two  foregoing  equations  gives 

req  =  ri  +  r2 (8) 

The  law  is  true  for  any  number  of  conductors  in  series;  it  may 
be  proved  by  successively  combining  them  into  groups  of  two. 

When  several  conductors  are  connected  in  parallel,  the  voltage 
across  them  is  common  to  all  the  branches,  so  that  we  have 


(9) 


where  i\,  12,  ...  are  the  currents  in  the  separate  branches.    The 
total  current  is  equal  to  the  sum  of  the  currents  in  the  separate 


8  THE  ELECTRIC  CIRCUIT  [ART.  3 

branches,  because  electricity  behaves  in  its  flow  like  an  incom- 
pressible fluid.  Thus,  the  equivalent  resistance,  req,  is  determined 
by  the  condition 

e  =  (i!  +  i,  +  etc.)  •  req (10) 

Substituting  the  values  of  ii,  iz,  etc.,  from  (9)  into  (10)  and 
canceling  e,  gives 

\/req  =  1/ri  +  l/r2  +  etc., (11) 

or,  in  words:  when  two  or  more  conductors  are  connected  in 
parallel,  the  reciprocal  of  the  equivalent  resistance  is  equal  to  the 
sum  of  the  reciprocals  of  the  individual  resistances. 

We  have  defined  conductance  as  the  reciprocal  of  resistance, 
so  that  eq.  (11)  may  be  written  also  in  the  form 

geq  =  gi  +  02  +  etc (12) 

It  will  thus  be  seen  that  it  is  convenient  to  use  conductances  in 
parallel  circuits  and  resistances  in  series  circuits.  The  simple 
rule  is:  Resistances  are  added  in  series;  conductances  are  added  in 
parallel.  This  rule  follows  directly  from  the  physical  concept  of 
resistance  and  conductance. 

Prob.  1.   Prove  that  when  two  conductors,  1  and  2,  are  in  parallel 

ti/i«  =  gi/gz  =  r2/n, (13) 

and  that  when  they  are  in  series 

ei/ci  =  ri/r«  =  QI/QI (14) 

Prob.  2.  Show  that  when  two  resistances  are  in  parallel  the  equivalent 
resistance 

req  =  rir2/(ri  +  r2), (15) 

and  that  for  two  conductances  in  series 

geq  =  0i0i/(0i  +  gj (16) 

Prob.  3.  Two  resistances,  TI  =  5  ohms  and  r2  =  7  ohms,  are  connected 
in  scries.  Resistance  ri  is  shunted  by  a  comparatively  high  resistance 
Ri  =  100  ohms;  r2  is  shunted  by  a  resistance  J?2  =  50  ohms.  What  is 
the  equivalent  resistance  of  the  whole  combination?  Solution: 

Equivalent  conductance  of  r,  and  Ri  is  0.2  +  0.01      =  0.21  mho; 
Equivalent  resistance  of  r,  and  Ri  is  1/0.21      =  4.76  ohms; 

Equivalent  conductance  of  r2  and  Ri  is 

0.1429  +  0.0200  =  0.1629  mho; 
Equivalent  resistance  of  r2  and  R«  is  1/0.1629  =  6.14  ohms. 

Ans.    4.76  +  6.14  =  10.90  ohms. 

Prob.  4.  Four  resistances,  r,  =  1.2,  r2  =  1.7,  R  =  25,  and  r0  = 
750  ohms,  are  connected  as  shown  in  Fig.  2.  The  generator  voltage 


CHAP.  I] 


DIRECT-CURRENT  CIRCUITS 


9 


between  the  points  A  and  B  is  500  volts.  Determine  the  current  through 
the  resistance  R  and  the  voltage  across  this  resistance.1  Solution:  Com- 
bine the  resistances  r^  and  R  into  one;  determine  the  conductance 
l/(R  +  r2),  and  combine  it  with  the  leakage  conductance  l/r0.  De- 


N  D 

FIG.  2.    A  series-parallel  combination  of  resistances. 


termine  the  equivalent  resistance  between  the  points  M  and  N,  and  the 
total  resistance  between  A  and  B.  Having  found  the  total  current,  sub- 
tract from  the  generator  voltage  the  voltage  drop  in  the  part  AM  of  the 
line.  This  will  give  the  voltage  across  M  N,  and  consequently  the  value 
of  the  leakage  current.  After  this, 
the  drop  in  rt  is  determined,  and 
thus  the  voltage  across  the  resist- 
ance R  is  found. 
Ans.  447.3  volts;  17.8  amps. 

Prob.  5.  The  armature  winding 
of  a  direct-current  machine  (Fig.  3) 
consists  of  108  coils;  the  conduct- 
ance of  each  coil  is  61  mhos.  The 
coils  are  connected  in  series  in  such 
a  way  that  the  circuit  is  closed 
upon  itself.  Two  positive  and  two 
negative  brushes  are  placed  alter- 
nately at  four  equidistant  points  of 
the  winding,  so  as  to  divide  it  into 
four  branches  in  parallel.  The  two 
positive  brushes  are  connected  to- 
gether, as  are  also  the  two  negative 
brushes.  What  is  the  equivalent 
resistance  of  the  armature  between 
the  terminals  of  the  machine? 

Ans.     0.1106  ohm.2 


Terminals      — 
FIG.  3.     A  four-pole  multiple  winding. 


1  This  combination  represents  a  transmission  line  the  resistance  of  which 
is  ri  +  raj  the  useful  load  resistance  is  represented  by  R,  and  the  leakage  re- 
sistance by  TO.    The  problem  in  a  generalized  form  is  of  great  importance  in 
the  theory  of  alternating-current  circuits  (see  Figs.  41  and  42). 

2  For  details  of  armature  windings  see  the  author's  Experimental  Electrical 
Engineering,  Vol.  2,  chap.  30. 


10  THE  ELECTRIC  CIRCUIT  [ART.  4 

Prob.  6.  Two  equal  resistances  of  r  ohms  each  are  connected  in 
series.  When  one  of  them  is  shunted  by  an  unknown  resistance  R,  the 
total  resistance  of  the  combination  decreases  by  10  per  cent.  Find  the 
value  of  R.  A™-  4r- 

4.  Electric  Power.  The  electric  power  (energy  per  unit 
time)  converted  into  heat  in  a  conductor  is  found  by  experi- 
ment to  be  proportional  to  the  resistance  of  the  conductor  and 
to  the  square  of  the  current  (Joule's  law).  The  practical  unit  of 
power,  the  watt,  is  so  selected  that  the  coefficient  of  proportion- 
ality is  unity,  or  the  power 

P    =    i2r    =    i2/g (17) 

Either  i  or  r  may  be  eliminated  from  this  expression,  using  Ohm's 
law.  This  gives  three  more  expressions  for  power: 

P  =  e.i  =  e2/r  =  etg (18) 

All  these  expressions  are  used  in  practice,  depending  upon  which 
quantities  are  known  in  a  particular  case. 

The  expression  e  •  i  is  the  fundamental  one;  it  is  analogous  to 
the  expression  h  •  q  for  the  power  lost  by  friction  in  a  pipe  in  which 
a  fluid  is  in  uniform  motion.  In  the  pipe,  the  energy  lost  per 
unit  time  is  equal  to  the  rate  of  discharge  q  times  the  head  h  lost 
in  friction;  in  other  words,  it  is  equal  to  the  quantity  factor 
times  the  intensity  factor.  In  an  electric  circuit  the  current  i 
is  the  quantity  factor,  while  the  voltage  drop  e  is  the  intensity 
factor. 

If  P  in  eqs.  (17)  and  (18)  is  expressed  in  kilowatts,  or  in  mega- 
watts, a  numerical  factor  equal  to  10~3  or  10~6  respectively  is 
introduced  on  the  right-hand  side  of  the  equation.  Sometimes 
the  output  of  a  motor  is  measured  in  horse-power;  the  English 
horse-power  is  equal  to  746  watts,  while  the  metric  horse-power 
is  736  watts.  It  is  strongly  recommended  by  the  International 
Electrotechnical  Commission  that  the  odd  and  superfluous  unit 
"horse-power"  be  dropped  altogether  and  that  mechanical  power 
be  expressed  also  in  watts  (or  kilowatts).  This  means  that  elec- 
tric motors  as  well  as  generators  should  be  rated  in  kilowatts. 

Sometimes  the  duty  of  a  machine  is  expressed  in  kilogram- 
meters  per  second;  the  conversion  ratio  to  watts  is: 

1  kg.-m.  per  second  =  9.806  watts. 


CHAP.  I]  DIRECT-CURRENT  CIRCUITS  11 

In  many  cases,  however,  it  is  not  necessary  to  introduce  either 
kilogram-meters  or  calories,  since  mechanical,  thermal,  and  electri- 
cal energy  can  all  be  expressed  in  joules  (watt-seconds). 

If  a  conductor  contains  a  local  e.m.f.  ei,  the  power  commu- 
nicated to  this  part  of  the  circuit,  between  its  terminals,  is 
equal  to  et  •  i,  where  et  is  the  terminal  voltage.  But  the  power 
i2r  converted  into  heat  may  be  either  smaller  or  larger  than  eti, 
depending  upon  the  polarity  or  direction  of  e\.  Multiplying  both 
sides  of  eq.  (4)  by  i,  we  find  that  the  power 

P  =  eti  +  eii  =  iV (19) 

Let  et  be  positive,  that  is,  in  the  same  direction  as  et  (for  instance, 
an  extra  battery  or  generator  connected  into  the  circuit  to  boost 
the  voltage);  the  power  fV  converted  into  heat  is  in  this  case 
larger  than  eti,  because  the  power  supplied  by  the  local  source 
of  e.m.f.  is  also  converted  into  heat.  If,  however,  ei  is  negative, 
that  is,  if  it  acts  as  a  counter-e.m.f.  (which  is  usually  the  case  in 
practice),  the  power  converted  into  heat  is  smaller  than  eti.  In 
this  case  the  power  eii  is  communicated  to  the  local  source  of 
e.m.f.  If  this  source  is  a  storage  battery,  the  energy  is  stored  in 
chemical  form,  and  may  be  made  available  at  a  later  time;  if  it 
is  a  motor,  the  energy  is  converted  into  mechanical  work  on  the 
motor  shaft.  Let,  for  instance,  the  voltage  at  the  terminals  of  a 
circuit  be  110  volts,  and  let  the  counter-e.m.f.  of  a  motor  in  the 
circuit  be  100  volts;  assume  the  current  through  the  circuit  to 
be  20  amp.  Then  the  voltage  drop  due  to  resistance  in  the 
conductors  is  only  10  volts,  and  the  power  converted  into  heat 
is  200  watts.  The  power  communicated  to  the  motor  is  2000 
watts,  and  the  total  power  supplied  to  the  circuit  is  2200  watts. 

The  unit  of  electrical  energy  is  the  watt-second,  or  joule.  When 
the  heat  dissipated  in  a  conductor  must  be  expressed  in  thermal 
units,  use  the  relation 

1  kg.-calorie  =  4186  joules. 

Prob.  1.  The  armature  current  of  a  220-volt  direct-current  motor  at 
a  certain  load  is  63  amp.,  and  the  armature  resistance  is  0.14  ohm.  How 
much  electric  power  is  converted  into  mechanical  form,  and  what  is  the 
torque  developed  by  the  armature  if  the  speed  is  1050  r.p.m.? 

Ans.     13.3  kw.;  12.3  m. -kg. 

Prob.  2.  If  the  currents  in  the  shunted  resistances  Ri  and  Rz  (problem 
3,  Art.  3)  represent  pure  loss  of  power,  what  is  the  efficiency  of  the  whole 
arrangement?  Solution:  Let  the  voltage  across  the  resistances  n  and 


12  THE  ELECTRIC  CIRCUIT  [ART.  4 

Ri  be  r.  Then  the  voltage  across  r2  and  Ri  is  e  •  (6.14/4.76)  =  1.29  e. 
Hence,  the  useful  power  is  e2/5  +  (1.29  <?)2/7  =  0.438  e2  watts.  The  power 
lost  in  the  resistances  fl,  and  R2  is  e2/100  +  (1.29  e)2/50  =  0.0433  e2  watts. 
The  efficiency  is  43.8/(43.8  +  4.33)  =  90  per  cent. 

Prob.  3.  The  heating  element  of  a  110- volt  electric  kettle  must  be 
designed  so  that  it  will  heat  1.5  liters  (1  liter  =  1  cu.  decimeter)  of  water 
at  a  rate  of  10°  C.  per  minute,  assuming  no  losses  by  radiation.  What 
are  the  resistance  of  the  element  and  the  rated  current  of  the  utensil? 

Ans.     11.6  ohms;  9.5  amp. 

Prob.  4.  It  is  required  to  calculate  the  exciting  current  i,  the  number 
of  turns  n  per  pole,  and  the  resistance  r  per  turn  of  a  field  coil  of  a  5000-kw. 
6-pole  turbo-alternator,  from  the  following  data:  The  excitation  required 
at  the  rated  load  is  9000  amp  .-turns  per  pole;  at  short  overloads  12,000 
amp.-turns  per  pole  are  needed.  The  external  area  of  the  field  coil  is 
280  sq.  dm.;  in  continuous  service,  4  sq.  cm.  of  cooling  surface  must  be 
allowed  per  watt  converted  into  heat,  in  order  to  avoid  overheating  the 
coils.  The  exciter  voltage  is  125,  and  during  the  overload  about  10  per 
cent  of  this  voltage  must  be  absorbed  in  the  field  rheostat,  as  a  margin. 
Hint:  Solve  the  following  three  equations;  in  =  9000;  i*rn  =  28,000/4; 
(12,000/?i)nr  =  0.9  X  125/6. 

Ans.     500  amp.-   18  turns;   1.562  X  10~3  ohms. 


CHAPTER  II 

FUNDAMENTAL  ELECTRICAL  RELATIONS  IN  DIRECT- 
CURRENT  CIRCUITS  — (Continued) 

5.  Resistivity  and  Conductivity.  A  cylindrical  conductor 
may  be  considered  as  a  combination  of  unit  conductors  in  series 
and  in  parallel.  For  instance,  a  wire  12  m.  long  and  having  a 
cross-section  of  70  sq.  mm.  may  be  regarded  as  composed  of  70  X 
12  =  840  unit  conductors,  each  of  one  square  millimeter  cross- 
section,  and  one  meter  long.  These  unit  conductors  are  first 
combined  into  sets  of  70  in  parallel,  and  then  the  12  sets  are 
connected  in  series.  The  resistance  of  such  a  unit  conductor, 
made  of  copper,  and  at  a  temperature  of  0°  C.,  is  about  0.016 
ohm.  A  set  of  70  unit  conductors  in  parallel  has  7V  of  the  resist- 
ance of  one,  because  the  current  is  offered  70  paths,  instead  of 
one;  twelve  sets  connected  in  series  offer  twelve  times  the  resist- 
ance of  one  set.  Therefore,  the  resistance  of  the  given  conduc- 
tor is  (0.016/70)  X  12  =  0.002743  ohm. 

Each  material  is  characterized  by  the  resistance  of  a  unit  con- 
ductor made  out  of  it.  The  resistance  of  a  unit  conductor  at  a 
specified  temperature  is  called  the  resistivity l  of  the  material  and 
is  denoted  by  p.  Thus,  the  resistance  of  a  conductor  of  a  length  I 
and  cross-section  A  is 

r  =  p-l/A (20) 

The  numerical  value  of  p  depends  upon  the  units  of  length  and 
resistance  used.  A  unit  conductor  may,  for  instance,  have  a 
cross-section  of  one  square  millimeter,  and  may  be  one  meter,  or 
one  kilometer  long;  or  it  may  be  a  centimeter  cube.  In  the 
English  system  it  may  have  a  cross-section  equal  to  one  circular 
mil,  and  a  length  of  one  foot,  one  thousand  feet,  one  mile,  or  any 
such  specified  length.  Besides,  the  resistivity  may  be  expressed  in 
ohms,  megohms,  or  microhms.  In  each  case,  the  unit  of  resistance 
and  the  units  in  which  the  dimensions  of  I  and  A  are  expressed  in 
1  The  older  name  is  "  specific  resistance." 
13 


14  THE  ELECTRIC  CIRCUIT  [ART.  5 

formula  (20)  are  selected  so  as  to  suit  the  convenience  of  the 
user  of  the  formula.  For  the  values  of  p  for  different  materials 
see  one  of  the  various  handbooks  and  pocketbooks  for  electrical 
engineers. 

In  some  cases  it  is  more  convenient  to  use  the  conductance  of 
the  unit  conductor,  instead  of  its  resistance.  The  conductance 
of  a  unit  conductor,  at  the  specified  temperature,  is  called  the 
electric  conductivity  (or  specific  conductance)  of  the  material;  the 
conductivity  is  the  reciprocal  of  the  resistivity  of  the  same'mate- 
rial.  Denoting  this  conductivity  by  7,  we  have  7  =  1/p.  By 
reasoning  similar  to  that  given  above,  we  find  that  the  conduc- 
tance of  a  conductor  having  the  dimensions  I  and  A  is 

g  =  7-A/l  ........     (21) 


Prob.  1.  The  resistivity  of  aluminum  equals  2.GG  microhms  per  cubic 
centimeter;  what  is  its  conductivity  per  mil-foot?  Solution:  The  re- 
sistance of  a  conductor  one  foot  long  and  having  a  cross-section  equal  to 
one  circular  mil  is  2.66  X  1CT6  X  197,300  X  30.48  =  16  ohms;  where 
197,300  =  (1000/2.54)2  X  4/V  is  the  factor  for  converting  square  centi- 
meters into  circular  mils,  and  30.48  is  the  number  of  centimeters  in  one 
foot.  Ans.  0.0625  mho  per  circular  mil-foot. 

Prob.  2.  Each  field  coil  of  an  electric  machine  has  720  turns,  the 
average  length  of  a  turn  being  about  1.5  m.  What  size  wire  is  required 
if  the  hot  resistance  of  the  coil  is  to  be  1.14  ohms?  According  to  the 
A.  I.  E.  E.  Standardization  Rules,  a  temperature  rise  of  50°  C.  is  allowed 
above  the  air  at  25°  C.  Ans.  About  20  sq.  mm. 

Prob.  3.  A  given  current  i  is  to  be  transmitted  at  a  given  voltage 
between  two  given  localities  whose  distance  apart  is  I.  Deduce  an 
expression  for  the  most  economical  size  of  the  line  conductor.  A  small 
conductor  means  a  saving  in  the  original  investment,  but  a  higher  operat- 
ing cost  on  account  of  the  power  lost  in  the  conductor,  and  vice  versa. 
The  most  economical  conductor  is  one  for  which  the  annual  interest  and 
depreciation  plus  the  cost  of  the  i*r  loss  per  year  is  a  minimum.  Solu- 
tion :  Let  the  cost  of  one  watt-year  be  p  cents,  and  let  the  conductor  cost 
q  cents  per  cubic  centimeter,  installed.  Let  5  be  the  annual  interest 
and  depreciation  in  per  cent  to  be  allowed  on  the  original  cost  of  the  con- 
ductor. The  cost  of  the  power  lost  in  the  line  is  pPpl/A,  and  the  initial 
cost  of  the  conductor  is  qlA.  The  condition  of  the  problem  is  that 

pVpl/A  +  sqlA  +K  =  mm.,  .....     (22) 

where  the  constant  K  represents  the  interest  and  depreciation  on  the 
poles,  insulators,  etc.,  the  size  of  which  is  essentially  independent  of  the 
size  of  the  conductor.  Equating  the  first  derivative  with  respect  to  A  to 
zero,  we  get  -pt'VA2  +  6q  =  0,  or 

pi*P/A  =  SqA  .....  .     (22a) 


CHAP.  II]  DIRECT-CURRENT  CIRCUITS  15 

In  other  words,  the  most  economical  cross-section  is  that  for  which  the 
sum  charged  to  the  annual  interest  and  depreciation  is  equal  to  the  cost 
of  the  wasted  energy.  This  result  is  independent  of  the  length  of  the 
line  and  of  the  voltage,  and  is  known  as  Kelvin's  law  of  economy.1  Know- 
ing all  the  data,  the  cross-section  A  can  be  calculated  from  condition 
(22a) ;  see  also  problems  6  and  7  in  Art.  6. 

Prob.  4.  A  transmission  line  from  the  generating  station  A  to  a  place 
B  is  I  kilometers  long.  At  B  the  line  is  divided  into  two  branches;  one 
to  C,  li  km.  long,  and  carrying  a  current  tij  the  other  to  D,  h  km.  long, 
and  eanying  a  current  t'i.  The  total  permissible  voltage  drop  from  A 
to  either  C  or  D  is  «  volts  (one  way).  Determine  the  sizes  of  the  con- 
ductors in  the  three  parts  of  the  line  so  as  to  make  the  total  initial  cost 
of  copper  a  minimum.  Solution:  Take  the  unknown  voltage  drop  x 
from  A  to  B  as  the  independent  variable;  then  the  three  cross-sections 
are  determined  by  the  conditions  ipl/A  =  x,  iipZi/Ai  =  «  —  x,  and 
*tpl*/At  =  f  —  x.  The  value  of  x  itself  is  determined  by  the  condition 
that  IA  +  M.I  +  ItAi  =  min.  Substituting  the  values  of  A,  Al  and  .4, 
into  this  expression,  and  equating  the  first  derivative  with  respect  to  x 
to  zero,  we  get  tP/x*  =  »*iii*/(e  -  *)*  +  uV/(«  -  *)*.  Extracting  the 
square  root  of  each  member  of  this  equation  and  solving  for  x,  we  find  that 


+  (»VO  (Wl* 

Having  found  x,  the  three  cross-sections  are  easily  calculated  from  the 
three  conditions  written  above. 

6.  Current  Density  and  Voltage  Gradient.  When  a  current 
is  distributed  uniformly  over  the  cross-section  of  a  cylindrical 
conductor,  it  is  convenient  to  speak  of  the  current  density,  or  the 
current  per  unit  cross-section  of  the  conductor.  Denoting  this 
density  by  t7,  we  have 

U  =  i/A (23) 

U  is  measured  hi  amperes  (or  kiloamperes,  milliamperes,  etc.) 
per  square  centimeter,  or  per  square  millimeter.  The  current 
density  is  numerically  equal  to  the  current  through  each  unit 
conductor  of  which  the  given  conductor  consists. 

1  In  practice  the  selection  of  the  cross-section  of  a  line  conductor  is 
determined  by  many  other  considerations  besides  that  of  economy,  as  out- 
lined above.  For  instance,  it  may  be  desired  to  reduce  the  original  invest- 
ment to  a  minimum  while  the  load  is  small,  and  to  change  the  conductors 
to  a  larger  size  afterwards.  The  problem  above  is  intended  only  to  introduce 
the  reader  into  this  subject.  He  will  find  numerous  contributions  treating 
of  more  complicated  cases  in  various  periodicals  and  transactions.  See  also 
A.  C.  Perrine,  Electrical  Conductors,  Chapter  8. 


16  THE  ELECTRIC  CIRCUIT  [ART.  6 

When  the  voltage  drop  is  distributed  uniformly  along  a  con- 
ductor, it  is  convenient  to  speak  of  voltage  drop  per  unit  length. 
This  voltage  drop  across  a  unit  length  is  called  the  voltage  gradient, 
and  is  measured  in  volts,  kilovolts,  millivolts,  etc.,  per  meter  or 
per  centimeter.  Denoting  the  voltage  gradient  by  G,  we  have 

G  =  e/l (24) 

The  value  of  G  characterizes  the  electrical  condition  at  a  point 
(or  a  cross-section)  of  the  conductor;  for  this  reason  G  is  some- 
times called  the  electric  intensity  at  a  point. 

Having  previously  introduced  the  resistance  p  and  the  conduct- 
ance 7  of  a  unit  conductor,  we  can  now  write  Ohm's  law  for  the 
unit  conductor,  in  the  form 

G=PU  =  V/y (25) 

Equation  (25)  has  a  definite  meaning  also  without  the  concept 
of  the  unit  conductor;  namely,  it  gives  the  relation  between  the 
voltage  gradient  and  the  current  density  at  a  point,  for  a  given 
material.  The  reader  can  easily  think  of  a  thermal  analogue. 
Hooke's  law  for  elastic  materials  is  also  somewhat  analogous  to 
eq.  (25),  because  it  expresses  a  straight-line  relation  between  the 
cause  and  the  effect.  Equation  (25)  can  be  deduced  directly 
from  eq.  (1)  by  writing  the  latter  in  the  form  Gl  =  (pl/A)  •  UA 
and  canceling  I  and  A. 

Prob.  1.  What  is  the  voltage  drop  per  kilometer  of  a  copper  wire 
having  a  cross-section  of  70  sq.  mm.,  and  carrying  a  current  of  150  amp.? 
Solution:  U=  150/70=  2.143  amp.  per  sq.  mm.  The  conductivity  of 
copper  7,  at  the  temperature  of  the  line,  is  equal  to  57  mhos  for  a  unit 
conductor  of  one  square  millimeter  cross-section  and  one  meter  long. 
Therefore,  the  electric  intensity  or  the  voltage  drop  per  meter  of  length, 
according  to  formula  (25),  is  2.143/57=  0.0376  volt  per  meter. 

Ans.    37.6  volts/km. 

Prob.  2.  What  is  the  expression  for  power  converted  into  heat  in  a 
unit  conductor?  Ans. 

P  =  G  •  U  =  U*P  =  C72/T  =  7  •  G* (25a) 

Prob.  3.  What  is  the  amount  of  power  lost  in  the  conductor  con- 
sidered in  problem  1? 

Ans.  (0.0376  X  2.143)  X  70  X  1000  =  5640  watts. 
Prob.  4.  The  space  available  on  the  frame  of  a  generator  for  a  rec- 
tangular field  coil  is  16  X  12  cm.  for  the  inside  dimensions,  and  28  X 
24  cm.  for  the  outside  dimensions;  the  limiting  height  is  15  cm.  What 
current  density  can  be  allowed  in  the  coil,  if  12  sq.  cm.  of  exposed  surface 
are  required  per  watt  loss,  in  order  that  the  temperature  of  the  coil  shall 


CHAP.  II]  DIRECT-CURRENT  CIRCUITS  17 

not  exceed  the  safe  limit?  The  space  factor  of  the  coil  is  0.55;  in  other 
words,  55  per  cent  of  the  gross  space  is  occupied  by  copper,  the  rest  being 
taken  by  the  air  spaces  and  the  insulation.  Solution:  The  exposed 
surface  is  2(28  +  24)  X  15  =  1560 sq.  cm.;  therefore,  130  watts  loss 
can  be  allowed  in  the  coil.  With  a  space  factor  of  0.55  the  useful  cross- 
section  of  copper  is  6  X  15  X  0.55  =  49.5  sq.  cm.;  the  average  length 
of  one  turn  equals  2(22  +  18)  =  80  cm.  Therefore,  the  coil  contains 
4950  X  0.8  =  3960  unit  conductors,  each  one  meter  long  and  one  square 
millimeter  in  cross-section.  The  permissible  loss  per  unit  conductor  is 
130/3960  =  0.0328  watt.  Hence,  according  to  the  answer  to  problem  2 
above,  U  =  A/0.0328  X  57  =  1.37  amp.  per  sq.  mm.  This  result  is 
independent  of  the  size  of  the  wire,  as  long  as  the  space  factor  remains 
approximately  the  same.  The  maximum  ampere-turns  are'137  X  49.5  = 
6780,  and,  for  a  constant  space-factor,  are  also  independent  of  the  size 
of  the  wire. 

Prob.  5.  What  are  the  size  of  wire  and  the  exciting  current  in  the  pre- 
ceding problem  if  the  voltage  drop  must  not  exceed  20  volts  at  80°  C.? 

Ans.    4.74  sq.  mm.;  6.5  amp. 

Prob.  6.  Referring  to  problem  3  in  Art.  5,  what  is  the  general  expres- 
sion for  the  most  economical  current  density?  Ans.  U  =  Vs<?/(pp). 

Prob.  7.  Referring  to  the  preceding  problem,  what  is  the  most  eco- 
nomical current  density  if  copper  costs  15  cents  per  pound,  the  annual 
interest  and  depreciation  is  taken  at  12  per  cent,  and  the  estimated  cost 
of  wasted  power  is  22  dollars  per  kilowatt-year? 

Ans.    0.95  amp.  per  sq.  mm.,  taking  p  at  25°  C. 

7.  Kirchhoff's  Laws.  Consider  an  arbitrary  network  of 
conductors  (Fig.  4),  with  sources  of  e.m.f.  connected  in  one  or 
more  places.  When  such  a  system  is  left  to  itself,  definite  cur- 
rents will  flow  through  the  conductors,  and  definite  differences  of 
potential  will  be  established  between  the  junction  points  of  the 
conductors.  Thus,  if  all  the  resistances  and  e.m.fs.  are  given,  it 
ought  to  be  possible  to  calculate  the  magnitude  and  direction  of 
all  the  currents.  The  distribution  of  the  currents  is  such  that 
two  conditions  are  satisfied: 

(1)  As  much  current  flows  toward  each  junction  as  from  it, 
because  electricity  behaves  like  an  incompressible  fluid.  For  any 
junction  this  is  expressed  mathematically  by  the  equation 

(26) 


in  which  all  the  currents  flowing  toward  the  junction  are  taken 
with  the  sign  plus,  all  those  flowing  away  from  it  with  the  sign 
minus,  or  vice  versa.  Thus,  for  instance,  at  the  point  C  let  the 
currents  flowing  toward  the  junction  be  20  and  30  amp.  respec- 


18 


THE  ELECTRIC  CIRCUIT 


[ART.  7 


lively,  and  one  of  the  currents  flowing  away  from  it  be  40  amp. 
Then  the  fourth  current  must  necessarily  be  10  amp.  flowing 
away  from  C,  because  20  +  30  -  40  -  10  =  0.  Equation  (26) 
is  called  Kirchhoff's  first  law. 

(2)  The  sum  of  the  terminal  voltages  along  any  closed  circuit 
in  the  network  is  equal  to  zero,  or 

2et  =  0 (27) 

Consider,  for  instance,  the  path  ABCDEFA,&nd  connect  a  zero- 
center  voltmeter  first  between  A  and  B,  then  between  B  and  C, 
and  so  on,  every  time  transferring  both  terminals,  so  that  one 


Fia.  4.    A  network  of  conductors,  illustrating  Kirchhoff's  laws. 

particular  terminal  of  the  instrument  always  leads  the  other. 
Consider  the  deflections  to  one  side  of  the  zero  point  as  positive, 
to  the  other  side  as  negative.  Equation  (27)  means  that  the 
algebraic  sum  of  these  readings  is  equal  to  zero.  The  reason  is 
as  follows:  The  reading  A-B  shows  how  much  higher  is  the 
potential  at  B  than  that  at  A ;  the  reading  B-C  shows  the  amount 
by  which  the  potential  at  C  is  higher  than  that  at  B.  Hence, 
the  sum  of  the  two  readings  indicates  the  difference  of  potential 
or  the  voltage  between  C  and  A.  Consequently,  the  sum  of  all 
the  readings  around  the  closed  circuit  indicates  the  difference 
of  potential  between  A  and  A,  which  difference  is  evidently 
zero,  no  matter  by  which  closed  path  the  original  point  has  been 
reached.  The  reader  may  again  resort  to  analogues  in  order  to 


CHAP.  II]  DIRECT-CURRENT  CIRCUITS  19 

see  this  law  more  clearly.  For  instance,  the  potentials  at  the 
joints  may  be  likened  to  temperatures,  and  the  voltmeter  readings 
to  differences  of  temperature.  Or  the  potentials  may  be  com- 
pared to  absolute  pressures  in  a  network  of  pipes,  and  the  voltages 
et  to  the  differences  of  pressure.  Again,  the  potentials  of  the 
points  A,  B,  C,  etc.,  are  analogous  to  the  altitudes  of  certain 
points,  say  above  the  sea  level,  while  the  voltages  correspond  to 
their  relative  elevations.  In  all  such  cases  the  sum  of  the  differ- 
ences around  a  closed  path  is  equal  to  zero. 

Equation  (27)  is  usually  written  in  a  somewhat  different 
form,  because  the  values  of  et  are  usually  not  known,  so  that  it 
is  desirable  to  express  them  through  the  given  electromotive 
forces  and  the  resistances  of  the  conductors.  The  general  expres- 
sion (4)  of  Ohm's  law  holds  for  each  conductor  in  the  network. 
Write  these  expressions  for  all  the  conductors  along  a  closed 
path,  and  add  them  together,  term  by  term.  The  sum  of  the 
e<'s  is  equal  to  zero,  according  to  eq.  (27),  so  that  the  result  is 

Sei  =  Sir (28) 

This  form  of  eq.  (27)  is  known  as  Kirchhoff's  second  law.  In 
this  equation  a  certain  direction  of  currents  and  voltages  must  be 
assumed  as  positive.  Let,  for  instance,  in  the  circuit  ABCDEFA 
the  clockwise  direction  be  taken  as  positive;  that  is,  all  the  cur- 
rents flowing  clockwise  are  to  be  considered  positive,  and  also 
all  the  e.m.fs.  which  tend  to  produce  currents  in  the  clockwise 
direction.  Let  e\  =  70  volts,  and  ez  =  50  volts,  and  let  the 
resistances  of  the  conductors  be  2,  3,  5,  4,  8  and  6  ohms  respec- 
tively. Let  all  the  currents  be  known  except  that  in  DE,  and  let 
them  be  10,  15,  15,  3  and  5  amp.  respectively,  the  directions  being 
those  shown  in  the  figure.  Denote  the  unknown  current  in  DE 
by  x,  and  assume  it  to  flow  in  the  clockwise  direction.  Equation 
(28)  then  becomes 

70-50  =  10X2+15X3-15X5  +  4z-f3X8  +  5X6, 

from  which  x  =  —  6  amp.  In  other  words,  the  current  in  DE  is 
equal  to  6  amp.  and  is  flowing  counter-clockwise. 

For  a  given  network  of  conductors  the  number  of  equations 
of  the  form  (26)  is  equal  to  the  number  of  junction  points  less  one, 
because  the  equation  for  the  last  point  can  be  obtained  by  com- 
bining the  other  equations.  The  number  of  equations  of  the 


20 


THE  ELECTRIC  CIRCUIT 


[ART.  7 


form  (28)  is  equal  to  the  number  of  independent  closed  paths  in 
the  network.  The  total  number  of  equations  of  both  kinds  is 
just  equal  to  the  number  of  unknown  currents,  so  that  these 
currents  can  be  determined  by  solving  the  simultaneous  equations. 

Prob.  1.  A  constant  c.m.f.  of  110  volts  is  maintained  at  the  generat- 
ing station,  and  power  is  transmitted  through  a  line  having  a  resistance 
of  0.5  ohm  to  two  devices  in  parallel,  viz.,  a  resistor  of  10  ohms,  and  a 
motor  the  internal  resistance  of  which  is  5  ohms.  Calculate  the  line 
current  (a)  when  the  motor  armature  is  blocked,  and  (b)  when  it  revolves 
at  such  a  speed  that  the  counter-e.m.f.  is  90  volts.. 

Ans.     28.7  amp.;   13.05  amp. 


FIG.  5.    An  unbalanced  Wheatstone  bridge. 

Prob.  2.   Write  the  equations  for  the  six   unknown   currents   in   a 
Wheatstone  bridge  (Fig.  5)  when  it  is  not  balanced. 

Ans.     {6  =  i2  4-  {4;    it  =  ii  +  ig;     i,  =  i4  4-  ig-    ibrb  4-  ilTl  4-  i2Ta  =  e- 

Prob.  3.  Show  that  the  preceding  six  equations  are  reduced  to  three 
when  the  galvanometer  circuit  is  open. 


Fia.  6.    A  leaky  electric  circuit  containing  a  counter-e.m.f.;  this  is  an 
analogue  to  the  magnetic  circuit  of  a  loaded  electric  machine. 

Prob.  4.  Two  sources  of  c.m.f.,  Cl  and  c»  (Fig.  6),  are  connected  to  act 

;  each  other,  e,  being  larger  than  e,.     The  internal  resistances  of 

»  sources  are  r,  and  r2  respectively;  the  main  external  resistance  is 

e  insulation  between  the  terminals  of  the  sources  of  e  m  f    is 


CHAP.  II]  DIRECT-CURRENT    CIRCUITS  21 

imperfect,  and  the  leakage  conductances  are  represented  by  gn  and  gi2 
respectively.  Write  Kirchhoffs  equations  for  the  unknown  currents 
(a)  when  there  is  no  leakage;  (b)  when  gi2  =  0;  (c)  when  gi{  =  0;  and 
(d)  when  both  leakages  are  present.1 

Prob.  5.  A  telegraph  line  with  ground  return  has  a  resistance  of  r' 
ohms  per  kilometer,  and  a  leakage  conductance  to  the  ground  of  g'  mhos 
per  kilometer.2  The  voltage  at  the  receiver  end  of  the  line  is  E2,  the 
receiver  current  I2.  What  are  the  values  of  the  voltage  and  of  the  current 
at  a  distance  of  s  kilometer  from  the  receiving  station?  Solution:  Con- 
sider an  infinitesimal  length  ds  of  the  line,  at  a  distance  s  from  the  receiv- 
ing end,  and  let  the  voltage  to  the  ground  at  this  point  be  e.  If  i  is  the 
line  current  at  the  same  point,  then  the  leakage  current  corresponding  to 
the  element  ds  of  the  line  is  di,  and  we  have,  according  to  Ohm's  law, 
di  =  eg'ds,  where  g'ds  is  the  leakage  conductance  through  the  element  ds 
of  the  line.  For  the  element  of  the  line  itself,  Ohm's  law  gives  de  =  ir'ds. 
Substituting  the  value  of  e  from  the  first  equation  into  the  second,  gives 
d2i/ds2  =  r'g'i,  or  i  is  such  a  function  of  s  that  its  second  derivative  is 
proportional  to  the  function  itself.  The  solution  of  this  differential 
equation  is  i  =  A1i~ms  +  A2e+ms,  where  m  =  ^r'g',  and  A\  and  A2 
are  the  constants  of  integration.  However,  in  our  case  it  is  preferable 
to  express  the  solution  through  hyperbolic  functions,  in  the  form  i  = 
Ci  cosh  ms  +  C2  sinh  ms,  where  m  =  v/rV  and  C\  and  C2  are  the  con- 
stants of  integration.  The  reader  can  check  this  solution  by  substitut- 
ing it  in  the  differential  equation.  The  constants  of  integration  are 
determined  from  the  given  conditions  at  the  receiver  end  of  the  line. 
Namely,  from  di  =  eg'dswefmde  =  (1/00  (di/ds)  =  (m/g1)  (Cisinhms  + 
C2  cosh  ms).  For  s  =  0,  e  =  E2  and  i  =  I2.  Consequently,  C\  =  I2; 
C2  =  E2g'/m. 

Prob.  6.  Referring  to  the  preceding  problem,  the  resistance  of  a 
telegraph  line  is  7  ohms  per  kilometer,  and  the  insulation  resistance  to 
the  ground  is  1.2  megohms  per  kilometer;  the  line  is  400  kilometers 
long.  A  relay  of  300  ohms  resistance  and  requiring  0. 12  ampere  to  operate 
it,  is  connected  between  the  receiver  end  of  the  line  and  the  ground. 
Calculate  the  required  current  and  battery  voltage  at  the  sending  station. 

Ans.  0.195  amp.;  445  volts. 

1  The  electric  circuit  shown  in  Fig.  6  is  of  importance  because  it  serves 
as  a  good  analogue  to  the  magnetic  circuit  in  a  loaded  machine.    The  electro- 
motive forces  ei  and  e2  correspond  to  the  magnetomotive  forces  of  the  field 
and  the  armature  respectively,  the  reluctances  of  the  parts  of  the  main  path 
being  represented  by  2  r,  n  and  r2,  while  the  leakage  permeances  correspond 
to  <7a  and  giz.    See  the  author's  Magnetic  Circuit,  the  latter  part  of  Art.  40, 
and  problem  13. 

2  Primed  symbols  are  used  in  this  book  and  in  the  Magnetic  Circuit  where 
quantities  refer  to  unit  length. 


CHAPTER  III 
CONDUCTORS   OF  VARIABLE  CROSS-SECTION1 

8.   Current  Density  and  Voltage  Gradient  at  a  Point.     When 

the  cross-section  of  a  conductor  varies  along  its  length  (Fig.  7), 
the  voltage  drop  per  unit  length  and  the  current  density  are  also 
variable.  In  places  like  MN,  where  the  cross-section  of  the 
conductor  is  comparatively  small,  the  resistance  per  unit  length 
is  correspondingly  large,  and  vice  versa.  Consequently,  the  volt- 
age gradient  and  the  current  density  are  also  larger  at  MN  than, 
for  example,  at  PQ.  Equations  (23)  and  (24)  give  in  this  case 
only  an  average  current  density  and  an  average  voltage  gradient 
over  the  conductor. 


Q 

FIG.  7.    A  couductor  of  variable  cross-section,  showing  the  stream  lines  and 
equipotential  surfaces. 

The  lines  traversing  the  diagram  (Fig.  7)  represent  stream 
lines  and  equipotential  surfaces.  The  stream  lines,  marked  with 
arrowheads,  represent  the  direction  of  the  electric  flow,  while  the 
equipotential  surfaces  are  perpendicular  to  them,  and  are  the  loci 
of  points  of  equal  potential.  The  distribution  is  analogous  to  that 

1  This  chapter  may  be  omitted  if  desired,  because  it  is  not  necessary  for 
the  understanding  of  the  following  chapters  on  alternating  currents.  The 
importance  of  this  chapter  lies  in  the  fact  that  the  treatment  is  analogous 
to  that  of  the  electrostatic  circuit,  and  therefore  it  greatly  facilitates  the  study 
of  the  latter.  This  chapter  may,  therefore,  be  conveniently  studied  before 
taking  up  Chapter  14.  The  treatment  is  also  analogous  to  that  used  in  the 
author's  Magnetic  Circuit. 


22 


CHAP.  Ill]    CONDUCTORS  OF  VARIABLE  CROSS-SECTION  23 

which  obtains  in  the  flow  of  heat;  the  stream  lines  indicate  the 
direction  of  the  flow  of  heat,  while  the  equipotential  surfaces  are 
analogous  to  those  of  equal  temperature. 

In  order  to  understand  the  meaning  of  equipotential  surfac.es, 
let  one  lead  of  a  voltmeter  be  applied  at  one  of  the  terminals  of 
the  conductor,  and  let  the  other  lead  be  moved  about  inside  the 
conductor  (assuming  this  to  be  possible),  marking  the  points  for 
which  the  deflection  of  the  voltmeter  remains  the  same.  All 
the  points  for  which  the  reading  is,  let  us  say,  10  volts  form  an 
equipotential  surface;  while  all  those  for  which  the  voltmeter 
reads  11  volts  form  another  equipotential  surface,  and  so  on. 
Between  two  points  on  the  same  equipotential  surface  the  volt- 
meter reading  is  evidently  zero.  The  equipotential  surfaces  are 
perpendicular  to  the  stream  lines,  because  if  there  were  a  com- 
ponent of  flow  along  an  equipotential  surface  there  would  be  an 
ir  drop  between  two  points  on  the  same  surface,  and  the  voltage 
between  these  two  points  could  not  be  zero. 

Stream  lines  and  equipotential  surfaces  give  a  clear  idea  of 
the  character  of  flow  of  a  current  in  a  conductor  of  irregular  shape, 
especially  if  they  are  drawn  to  correspond  to  equal  increments  of 
current  and  voltage.  This  means  that  the  lines  of  flow  should 
be  drawn  so  as  to  define  tubes  of  current  of  equal  strength.  For 
instance,  in  Fig.  7  the  current  included  between  any  two  adjacent 
stream  lines  is  supposed  to  be  the  same — let  us  say,  equal  to  one 
ampere.  Similarly,  the  voltage  between  any  two  adjacent  equi- 
potential surfaces  should  be  the  same;  for  example,  one  volt.  If 
the  lines  are  drawn  sufficiently  close  together,  they  give  complete 
information  about  the  voltage  and  current  relations  in  the  differ- 
ent parts  of  the  conductor,  and  also  show  places  of  high  and  low 
current  density  and  voltage  gradient. 

The  true  current  density  at  a  point  is  obtained  by  considering 
an  infinitesimal  tube  of  current  di  and  dividing  di  by  the  infini- 
tesimal cross-section  dA  of  the  tube  at  the  point  under  consider- 
ation. Then,  instead  of  eq.  (23),  we  have 

U=di/dA (29) 

If,  on  the  other  hand,  it  is  desired  to  express  the  total  current 
through  the  density,  the  preceding  relation  gives 


i  =  /     UdA, (30) 


24  THE  ELECTRIC  CIRCUIT  [ART.  8 

the  integration  to  be  extended  over  the  whole  equipotential  sur- 
face, U  being  a  function  of  the  position  of  dA.  In  other  words, 
current  is  the  surface  integral  of  current  density. 

..  The  relation  between  the  variable  intensity  G  along  the  conduc- 
tor, and  the  total  voltage  e  at  its  terminals,  is  no  longer  expressed 
by  the  simple  relation  (24),  applicable  to  the  whole  conductor. 
Relation  (24)  must  now  be  written  for  an  infinitesimal  length  dl 
of  a  stream  line,  because  G  is  constant  only  for  an  infinitesimal 
length.  The  definition  of  G  remains  the  same,  namely,  G  is  the 
rate  of  variation  of  voltage  per  unit  length  of  the  conductor. 
Thus,  denoting  by  de  the  voltage  between  two  adjacent  equi- 
potential surfaces  at  a  distance  dl  apart,  we  have 

G  =  de/dl,    or    de  =  G>dl (31) 

The  total  voltage  e  between  the  terminals  of  the  conductor  is 
equal  to  the  sum  of  these  infinitesimal  drops,  or 


f 

Jo 


G-dl (32) 


Equation  (32)  is  expressed  in  words  by  saying  that  voltage  is  the 
line  integral  of  electric  intensity  (or  voltage  gradient). 

A  clear  understanding  of  relations  (31).  and  (32)  is  of  para- 
mount importance  in  the  study  of  electrostatic  and  magnetic 
phenomena.  This  will  be  aided  by  recalling  to  mind  the  thermal 
analogy  previously  used.  In  the  case  of  the  flow  of  heat,  G 
corresponds  to  the  rate  of  change  in  temperature  per  unit  length 
of  the  rod,  while  e  represents  the  total  difference  of  temperature 
between  the  ends  of  the  rod.  Equation  (31)  expresses  the  fact 
that,  by  taking  the  rate  at  a  certain  point  and  multiplying  it  by 
a  very  short  element  of  the  length  of  the  rod,  the  actual  difference 
of  temperature  between  the  ends  of  this  element  is  obtained. 
Thus,  for  instance,  let  the  drop  in  temperature  at  some  point 
of  the  rod  be  equal  to  2.5°  C.  per  meter  length.  Then  the  actual 
drop  in  a  very  short  element,  say  0.1  mm.,  is  2.5  X  0.0001  = 
0.00025°  C.  The  element  of  length  must  be  small,  because  by 
supposition  the  cross-section  of  the  rod  is  not  constant,  and  the 
rate  of  drop  is  consequently  variable.  For  a  short  length  the  vari- 
able quantities  can  be  assumed  constant,  or,  more  correctly,  aver- 
age values  can  be  used.  Equation  (32)  thus  states  that  the  total 
difference  of  temperature  between  the  ends  of  the  rod  is  equal  to 
the  sum  (or  the  integral)  of  the  drops  in  the  very  small  elements. 


CHAP.  Ill]    CONDUCTORS  OF  VARIABLE  CROSS-SECTION  25 

Similarly,  in  a  pipe  of  variable  cross-section  the  rate  of  loss 
of  head  per  unit  length  is  variable,  so  that  it  is  only  possible  to 
speak  of  this  rate  G  at  a  point.  The  total  loss  of  pressure,  or 
head  e,  is  obtained  by  summing  up  the  small  losses  of  head  in 
infinitesimal  elements  of  the  pipe.  The  loss  of  pressure  for  a 
length  dl  isGdl;  the  total  head  e  is  the  integral  of  this  expression, 
over  the  whole  length  of  the  pipe.  This  is  expressed  mathemati- 
cally by  eq.  (32). 

Relation  (25)  between  G  and  U  holds  true  for  a  non-uniform 
flow  as  well,  because  it  merely  gives  a  relation  between  the  cause 
G  and  the  effect  U  at  a  point,  depending  only  upon  the  property 
of  the  material,  as  expressed  by  the  factor  7  or  p.  This  relation 
may  be  also  considered  as  Ohm's  law  for  an  infinitesimal  cylindri- 
cal conductor  of  length  dl  and  cross-section  dA,  namely, 

Gdl  =  de=  (Pdl/dA}UdA. 
Canceling  dA  and  dl,  relation  (25)  is  obtained. 

Prob.  1.  A  current  of  50  amp.  is  flowing  along  a  cylindrical  con- 
ductor 3  cm.  in  diameter.  The  resistivity  of  the  material  varies  in  con- 
centric layers  in  such  a  way  that  the  current  density  is  proportional  to 
the  cube  of  the  distance  from  the  axis.  What  is  the  current  density  at 
the  periphery?  Ans.  17.7  amp.  per  sq.  cm. 

Prob.  2.  A  conductor  of  circular  cross-section,  225  cm.  long,  has  the 
form  of  a  truncated  cone,  the  diameters  of  the  two  terminal  cross-sections 
being  1.2  cm.  and  3  cm.  respectively.  The  total  drop  at  a  certain  current 
is  65  volts.  What  is  the  general  expression  for  the  voltage  gradient  Gx 
at  a  distance  x  from  the  smaller  end? 

Ans.  Gx/G0  =  [a/(a  +  x)]2,  where  a  =  150  cm.  is  the  distance  from 
the  smaller  end  to  the  apex  of  the  cone,  and  G0=  0.723  volt  per  centimeter 
is  the  voltage  gradient  at  the  smaller  end.  G0  is  determined  from  eq. 

(32),  namely,  65  -  G0  f™  a*dx/(a  +  z)2. 

Jo 

Prob.  3.  A  non-linear  irregular  conductor,  made  of  homogeneous  ma- 
terial, has  a  current  density  U  and  an  electric  intensity  G,  varying  from 
point  to  point  in  magnitude  and  direction.  What  is  the  general  expres- 
sion for  the  power  converted  into  heat? 

Ans.     According  to  eq.  (25a), 

/\    f*  r 

y  J  J 

where  dv  is  the  element  of  volume  to  which  G  and  U  refer,  and  the  inte- 
gration is  extended  over  the  whole  volume  of  the  conductor.  The 
volume  dv  must  be  taken  as  a  cylinder  or  parallelepiped,  the  length  of 
which  is  in  the  direction  of  flow  of  the  current,  the  cross-section  being 
perpendicular  to  this  flow. 


26 


THE  ELECTRIC  CIRCUIT 


[ART. 


9.  The  Radial  Flow  of  Current.  The  solution  of  problems 
involving  a  non-uniform  flow  of  current  usually  requires  con- 
siderable facility  in  the  use  of  higher 
mathematics  beyond  ordinary  calculus. 
An  exception  to  this  statement  is  the 
simple  case  of  radial  flow  (Fig.  8)  be- 
tween two  concentric  electrodes,  cyl- 
indrical or  spherical.  The  following 
exercises  give  an  opportunity  for  prac- 
tice in  the  solution  of  problems  of  this 
kind.  They  serve  to  illustrate  the 
concepts  of  current  density  and  voltage 
gradient,  and  to  prepare  the  student's 
mind  for  the  solution  of  certain  prob- 
lems on  concentric  cables,  involving 
the  dielectric  and  magnetic  circuits. 


FIG.  8.  Flow  of  current  be- 
tween two  concentric  elec- 
trodes. 


Prob.  1.  Calculate  the  resistance  of  a  cylindrical  layer  of  mercury 
MM  (Fig.  8)  of  height  h  =  5  cm.,  between  two  concentric  cylindrical 
terminals  7\  and  T2,  the  radii  of  the  contact  surfaces  being  a  =  10  cm. 
and  b  =  18  cm.  The  resistivity  of  mercury  is  95  microhms  per  cubic  cen- 
timeter. Solution:  Take  an  infinitesimal  layer  of  the  mercury,  between 
the  radii  x  and  x  +  dx;  the  resistance  of  this  layer  is  P  •  dx/(2irxh). 
The  resistances  of  all  the  infinitesimal  concentric  layers  are  in  series; 
therefore,  r  is  obtained  by  integrating  the  foregoing  expression  between 
the  limits  a  and  6,  the  result  being 

r  =  [P/(2^)].Ln(6/a) (34) 

Ans.     r  =  1.775  microhms. 

Prob.  2.  In  the  preceding  problem,  when  a  current  of  10,000  amp. 
flows  through  the  mercury,  what  is  the  amount  of  heat  generated  per 
second  per  cubic  centimeter  of  mercury,  at  both  electrodes?  Solution: 
The  current  density  at  the  inner  electrode  is  10,000/(2x  X  10  X  5)  = 
31.8  amp.  per  square  centimeter.  According  to  eq.  (25a),  the  loss  of 
power  is  (31.8)2  X  95  X  10~«  =  0.0958  watt  per  cubic  centimeter.  The 
heat  loss  at  the  outer  electrode  is  0.096  X  (10/18)2  =  0.0295  watt  per 
cubic  centimeter. 

Prob.  3.  What  is  the  curve  of  electric  intensity  G  as  a  function  of  x 
in  the  preceding  two  problems,  and  what  are  the  limiting  values  of  Gl 

Ans.    An  equilateral  hyperbola;  Gi  =  3.02,  (72  =  1.677  millivolts  per 
centimeter  length. 

Prob.  4.  A  lead-covered  cable,  consisting  of  a  solid  circular  conductor 
of  A  square  millimeters  in  cross-section,  is  insulated  with  a  layer  of 
rubber  c  mm.  thick,  between  the  conductor  and  the  sheathing.  What  is 


CHAP.  Ill]     CONDUCTORS  OF  VARIABLE  CROSS-SECTION  27 

the  insulation  resistance  of  I  kilometers  of  such  a  cable,  if  the  resistivity 
of  rubber  is  p  megohms  per  centimeter  cube? 

Ans.  [P  X  10~*/(2  nt)]  •  Ln  (1  + 1-772  c/V.l)  megohms,  according  to 
eq.  (34). 

Prob.  6.  Show  that  by  doubling  the  thickness  of  the  insulation  in 
the  preceding  problem,  the  insulation  resistance  is  increased  less  than 
twice. 

Prob.  6.  A  current  is  flowing  through  a  hemispherical  shell  of  metal 
along  radial  lines.  Express  the  resistance  of  the  shell  as  a  function  of  its 
radii  a  and  6,  and  the  conductivity  y  of  the  material. 

Ans.     (&-o)/(2irya&). 

Prob.  7.  Apply  the  method  of  superposition  and  the  result  obtained 
in  Arts.  60  and  63,  to  the  calculation  of  the  resistance  of  an  unlimited 
conducting  medium  between  two  parallel  cylindrical  terminals.  Such  a 
case  obtains,  for  instance,  when  a  load  resistor  consists  of  two  vertical 
pipes  in  a  pond,  the  pipes  being  used  as  the  terminals,  the  current 
flowing  through  the  water. 

10.  The  Resistance  and  Conductance  of  Irregular  Paths. 
Let  a  conductor  of  irregular  shape  (Fig.  7)  be  connected  to  a 
source  of  constant  voltage  e.  The  power  converted  into  heat  in 
the  conductor  is  e2/r,  where  r  is  the  resistance  of  the  conductor. 
This  resistance  depends  upon  the  distribution  of  the  current  in 
the  body  of  the  conductor.  The  general  law,  demonstrated  by 
all  experiments,  is  that  the  distribution  of  the  current  is  such  as 
to  make  the  dissipated  energy  a  maximum.  Since  by  supposi- 
tion e  is  constant  (unlimited  supply),  the  resistance  r  must  be  a 
minimum. 

Let  now  the  same  conductor  be  connected  to  a  source  of 
constant  current  —  for  instance,  an  arc-light  machine.  The  dis- 
tribution of  the  current  in  the  conductor  is  such  as  to  effect  its 
passage  with  a  minimum  expenditure  of  energy,  that  is,  minimum 
voltage  at  the  terminals,  or  minimum  iV.  This  again  means 
that  the  resistance  r  is  a  minimum.  The  student  is  advised  to 
consider  similar  cases  in  the  flow  of  heat  or  of  a  fluid,  in  order  to 
make  the  matter  perfectly  clear  to  himself. 

The  general  law  of  nature  —  that  of  minimum  effort  or  mini- 
mum resistance  —  applies  in  all  such  cases,  and  is  used  in  the  cal- 
culation of  the  resistance  of  conductors  of  irregular  form.  The 
conductor  is  divided  into  small  parts  by  means  of  stream  lines 
and  equipotential  surfaces  as  shown  in  Fig.  7,  drawing  them  to  the 
best  of  one's  judgment.  These  small  cells  are  nearly  cylindrical 
in  form,  so  that  their  resistances  or  conductances  are  easily  esti- 


28  THE  ELECTRIC  CIRCUIT  [Anx.  11 

mated  by  using  their  mean  lengths  and  average  cross-sections. 
The  resistance  of  the  whole  conductor  is  found  by  properly  com- 
bining the  resistances  of  these  cells  in  series,  and  the  conductances 
of  the  filaments  thus  obtained  in  parallel.  Then  the  assumed 
shapes  of  the  stream  lines  and  of  the  equipotential  surfaces  are 
somewhat  modified,  and  the  resistance  is  calculated  again,  and 
so  on.  Thus,  by  successive  trials,  the  minimum  resistance,  or  the 
maximum  conductance,  of  the  given  conductor  is  found,  and  this 
is  the  true  value  of  resistance  or  conductance,  as  the  case  may 
be.  The  lines  corresponding  to  this  minimum  give  the  true  distri- 
bution of  currents  and  voltages  within  the  conductor. 

The  work  of  the  trials  is  made  more  systematic  by  following 
a  procedure  suggested  by  Lord  Rayleigh,  and  further  developed 
by  Dr.  Lehmann.  This  method  is  described  in  detail  in  Art.  54 
below,  in  application  to  the  electrostatic  field,  and  also  in  Art.  41 
of  the  author's  Magnetic  Circuit,  in  application  to  the  magnetic 
field.  The  student  will  have  no  difficulty  in  applying  the  method 
to  an  electro-conducting  circuit.  The  best  way  to  make  it  clear 
to  one's  self  is  actually  to  draw  a  conductor  of  irregular  shape 
'(in  two  dimensions  for  the  sake  of  simplicity)  and  to  calculate  its 
resistance  in  the  above-mentioned  manner.1 

11.  The  Law  of  Current  Refraction.  The  method  out- 
lined above  for  the  mapping  out  of  stream  lines  and  equipoten- 
tial surfaces  applies  only  in  a  homogeneous  conductor.  When  a 
current  passes  from  one  substance  to  another  (Fig.  9),  the  stream 
lines  suddenly  change  their  direction  at  the  dividing  surface  AB 
between  the  media,  and  in  so  doing  they  obey  the  law  of  cur- 
rent refraction,  which  is 

tan  0i/tan  02  =  71/72 (35) 

Here  0i  and  02  are  the  angles  of  incidence  and  refraction,  while 
71  and  72  are  the  respective  conductivities  of  the  two  media. 
This  equation  shows  that  the  lower  the  conductivity  of  a  sub- 
stance, the  more  nearly  do  the  stream  lines  approach  the  direc- 
1  In  two-dimensional  problems  of  this  kind,  the  properties  of  conjugate 
functions  may  be  used  when  the  geometric  forms  involved  can  be  expressed 
by  analytic  equations.  However,  the  purely  mathematical  difficulties  are 
such  as  to  make  this  method  applicable  only  in  a  comparatively  few  simple 
X  Maxwell,  Electricity  and  Magnetism,  Vol.  1,  p.  284;  J.  J. 
Thomson,  Recent  Researches  in  Electricity  and  Magnetism,  chap.  3;  Horace 
Lamb,  Hydrodynamics,  chap.  4. 


CHAP.  Ill]    CONDUCTORS  OF  VARIABLE  CROSS-SECTION 


29 


tion  of  the  normal  NiN*  at  the  dividing  surface.  In  this  way, 
the  path  between  two  given  points  is  shortened  in  the  medium 
of  lower,  and  is  lengthened  in  that  of  higher  conductivity,  by  such 
an  amount  in  each  case  that  the  total  conductance  of  the  composite 
conductor  is  larger  with  refraction  than  without  it.  Hence,  the 
existence  of  refraction  is  a  necessary  consequence  of  the  general 
law  of  least  resistance. 


Mediums, 
of  low  conductivity 


Medium  1, 
of  high  conductivity 


FIG.  9.     The  refraction  of  a  current,  or  of  a  flux. 


To  deduce  eq.  (35),  consider  a  tube  of  current  between  the 
equipotential  surfaces  ab  and  cd,  and  let  the  width  of  the  path  in 
the  direction  perpendicular  to  the  plane  of  the  paper  be  one 
centimeter.  Let  C/i  and  Uz  be  the  current  densities  in  the  tube, 
and  let  G\  and  Gz  be  the  corresponding  voltage  gradients.  Two 
conditions  must  be  satisfied,  namely,  (1)  the  total  current  through 
cd  is  equal  to  that  through  ab,  and  (2)  the  voltage  drop  along  ac 
is  the  same  as  that  along  bd.  These  conditions  are  expressed  by 
the  equations 

Ui-ab  =  Uz-cd 
and 

d  .  bd  =  (?2  •  ac. 

Dividing  the  first  equation  by  the  second  and  rearranging  the 
terms  gives 

Ui/Oi  _  Ut/Gt 

bd/ab       ac/cd 


30  THE  ELECTRIC  CIRCUIT  [ART.  11 

But.  according  to  eq.  (25),_l/j/6ri  =  71,  and  Ut/Gt  =  72-  From 
Fig.  9,  bd/ab  =  tan  0i,  and  ac/cd  =  tan  02.  By  substituting  these 
values  in  the  preceding  equation,  relation  (35)  is  obtained. 

Thus,  in  mapping  out  an  electro-conducting  circuit  in  two 
media,  the  stream  lines  must  be  so  drawn  as  to  satisfy  eq.  (35), 
and  the  conductance  must  be  a  maximum  for  the  combination, 
and  not  for  each  part  separately.  A  similar  law  applies  to  elec- 
trostatic and  magnetic  fluxes  (see  Art.  55  below,  and  Art.  41a  of 
the  author's  Magnetic  Circuit). 

Prob.  1.  Make  clear  to  yourself  the  reason  why  the  refraction  of 
light  follows  a  sine  law,  while  in  the  case  of  the  electric  current  it  is  a 
law  of  tangents. 

Prob.  2.  Show  that  total  refraction  is  impossible  for  an  electric  cur- 
rent. 

Prob.  3.  Draw  a  set  of  curves  giving  values  of  0X  for  different  values 
of  Ot  when  the  ratio  of  conductivities  is  1,  2,  10  and  100. 


CHAPTER  IV 

REPRESENTATION  OF  ALTERNATING  CURRENTS  AND 
VOLTAGES  BY  SINE-WAVES  AND  BY  VECTORS 

12.  Sinusoidal  Voltages  and  Currents.  A  large  proportion 
of  the  electric  power  used  for  lighting,  industrial  purposes,  and 
traction  is  generated  in  the  form  of  alternating  currents.  Some 
of  the  advantages  of  the  alternating  current  over  the  direct  cur- 
rent are:  (1)  Alternating-current  power  can  be  easily  converted 
into  power  at  a  higher  or  at  a  lower  voltage,  thus  making  possible 
the  transmission  of  power  over  long  distances;  (2)  the  genera- 
tion of  alternating  currents  is  simpler  than  that  of  direct  currents, 
the  latter  requiring  a  commutator,1  which  needs  constant  atten- 
tion in  operation;  and  (3),  by  combining  two  or  three  alternating- 
current  circuits  into  a  polyphase  system  it  is  possible  to  convert 
electric  into  mechanical  power,  using  motors  of  simple  and  rugged 
construction  (induction  motors  and  synchronous  motors). 

Alternating  voltage  waves  generated  by  commercial  alter- 
nators are  more  or  less  irregular  in  shape,  but  for  most  engineer- 
ing calculations  it  is  accurate  enough  to  assume  them  to  vary 
with  the  time  according  to  the  sine  law  (Fig.  10).  This  assump- 
tion simplifies  the  theory  and  calculations  greatly;  moreover, 
the  results  obtained  with  this  assumption  are  comparable  with 
one  another,  because  they  all  refer  to  a  standard  shape  of  the 
voltage  and  current  curves,  instead  of  a  particular  form  in  each 
specific  problem.  If  the  curve  of  a  voltage  or  current  differs 
greatly  from  the  sine-wave,  it  can  be  resolved  into  a  series  of  sine- 
waves  of  different  frequencies,  so  that  even  then  the  sine-wave 
remains  the  fundamental  form  (see  Art.  15  below).  Fig.  10 
shows  the  well-known  construction  of  a  sine-wave,  the  instan- 
taneous values  of  the  current  or  voltage  being  represented  as 

1  The  homopolar  machine,  which  is  a  direct-current  machine  without  a 
commutator,  has  not  proven,  up  to  the  present  time,  to  be  commercially  suc- 
cessful. 

31 


32 


THE  ELECTRIC  CIRCUIT 


[ART.  12 


ordinates,  against  time  as  abscissae.  Instead  of  actual  time  in 
seconds,  the  curve  is  sometimes  plotted  against  some  other  quan- 
tity proportional  to  time  —  for  instance,  fractions  of  a  complete 
cycle.  It  is  sometimes  convenient  to  use  as  abscissae  the  angu- 
lar positions  of  a  field  pole  of  the  alternator  with  respect  to  an 
armature  conductor  in  which  the  electromotive  force  under,  con- 
sideration is  induced. 


Fia.  10.     An  alternating  current  represented  by  a  sine-wave. 

To  construct  the  curve  of  an  alternating  current  or  voltage, 
draw  a  circle  the  radius  of  which  equals  the  maximum  value  of 
the  wave.  Divide  the  circle  into  a  certain  number  of  equal  or  un- 
equal parts,  such  as  ab,  be,  etc.,  and  mark  on  the  axis  of  abscissae 
points  a',  b',  c',  etc.,  corresponding  to  the  points  of  division  on 
the  circle.  That  is,  a'b'  is  either  equal  or  proportional  to  ab; 
b'c'  is  either  equal  or  proportional  to  be,  and  so  on.  In  general, 
an  abscissa  such  as  a'c'  represents,  to  a  certain  selected  scale,  the 
central  angle  u  corresponding  to  the  arc  ac.  The  lengtlTa'ra' 
represents  to  the  same  scale  an  angle  of  360  degrees,  or  the  time 
of  one  complete  cycle  of  the  wave.  The  ordinates  of  the  sine- 
wave  are  equal  to  the  corresponding  ordinates  of  the  circle.  For 
example,  the  point  c'"  on  the  curve  is  obtained  by  transferring 
the  ordinate  cc"  of  the  circle  to  the  corresponding  abscissa  a'c'. 

The  name  "sine-wave"  is  derived  from  the  fact  that  these 
ordinates  are  proportional  to  the  sines  of  the  abscissae,  which 
represent  to  some  scale  the  central  angles  of  the  circle  of  refer- 
ence. The  equation  of  the  curve  expresses  this  property  ana- 
lytically. Let  the  maximum  value  of  the  current,  which  is  also 


CHAP.  IV]  SINE-WAVES  AND  VECTORS  33 

equal  to  the  radius  of  the  circle,  be  denoted  by  7TO;  we  have  then 
from  the  triangle  Occ" 

.......     (36) 


where  the  ordinate  i  =  cc"  =  c'c'"  represents  the  instantaneous 
value  of  the  alternating  current,  at  the  moment  of  time  corre- 
sponding to  the  angle  u.  The  variable  angle  u  is  proportional 
to  the  time,  because  the  radius  Oc  which  generates  the  sine-wave 
is  assumed  to  revolve  at  a  uniform  speed.  Let  time  t  be  counted 
from  the  position  Oa  of  this  radius,  and  let  T  =a'mf  be  the  in- 
terval of  time  necessary  to  complete  one  revolution  of  the  radius, 
or  the  time  of  one  complete  cycle  of  the  alternating  wave.  When 
t  =  0,  u  =  0;  and  when  t  =  T,  u  =  2ir.  Therefore,  in  general, 

u  =  2irt/T,      .......     (37) 

because  this  expression  satisfies  the  foregoing  conditions.  Sub- 
stituting this  value  of  u  into  eq.  (36),  we  obtain 

i  =  Imsm(2irt/T)  .......     (38) 

For  the  values  of  t  =  0,  £  T,  T,  f  T,  etc.,  i  =  0,  as  one  would 
expect,  because  at  these  moments  the  current  changes  from 
positive  to  negative  values,  or  vice  versa.  At  t  =  j  T,  f  T,  £  T, 
etc.,  we  have  i  =  ±7m;  at  these  moments  the  current  reaches 
its  positive  and  negative  maxima.  Equation  (36)  is  used  when 
the  sine-wave  is  plotted  against  the  values  of  angle  as  abscissae. 
Equation  (38)  gives  the  same  curve  referred  to  time  as  abscissae. 
'in  practice,  the  rapidity  with  which  currents  and  voltages 
alternate  is  not  denoted  by  the  fraction  of  a  second  T  during 
which  a  cycle  is  completed,  but,  in  a  more  convenient  manner, 
by  the  number  of  cycles  per  second.  Thus,  instead  of  saying 
that  an  alternator  generates  current  which  completes  a  cycle 
within  ^  of  a  second,  it  is  customary  to  say  that  the  frequency 
of  the  current  is  60  cycles  per  second.  Denoting  the  frequency 
in  cycles  per  second  by  /,  we  have 

/  =  1/T,  ........     (39) 

and  consequently 

i  =  /msin2ir/«  ........     (40) 

This  is  the  usual  expression  for  an  alternating  current  having  a 
frequency  of  /  periods  per  second.  Analogously,  for  an  alternat- 
ing voltage  we  have 

e  =  Emsm2irft,  .......     (41) 


34  THE  ELECTRIC  CIRCUIT  [ART.  12 

where  Em  is  the  maximum  instantaneous  value,  also  called  the 
amplitude,  and  e  is  the  instantaneous  value  of  the  voltage  at  the 
time  t. 

In  numerical  calculations,  and  when  drawing  sine-waves,  the 
values  of  the  ordinates  for  various  values  of  u  or  t  are  obtained 
either  graphically,  as  in  Fig.  10,  or  from  a  table  of  sines.  For 
approximate  calculations,  values  of  sines  can  be  taken  from  a 
slide-rule.  In  the  problems  which  follow,  the  student  is  advised 
to  become  familiar  with  each  of  the  three  methods  of  obtaining 
values  of  sines. 

In  some  cases  one  has  to  deal  with  two  currents  or  voltages 
of  the  same  frequency  —  for  instance,  in  two  different  parts  of  the 
same  circuit.  The  two  corresponding  sine- waves  (Fig.  11)  usu- 


FIG.  11.    Two  alternating  currents  displaced  in  phase  by  an  angle  <f>. 

ally  differ  in  amplitude,  and  also  pass  through  zero  at  different 
instants.  Thus,  in  Fig.  11,  when  the  current  1  is  at  a  maximum, 
the  current  2  is  still  growing,  and  passes  through  its  maximum 
somewhat  later.  In  other  words,  the  current  2  lags  behind  the 
current  1,  or,  what  is  the  same,  the  current  1  leads  the  cur- 
rent 2.  The  angle  0  between  the  zero  points  (or  between  the 
maxima)  of  the  two  waves  is  called  the  angle  of  phase  differ- 
ence, or  simply  the  phase  angle.  If  we  regard  the  two  waves 
as  being  formed  by  the  revolving  radii  Im'  and  7TO",  then  </>  is  the 
angle  between  the  radii  at  any  instant. 

When  the  two  waves  are  of  different  frequencies,  there  is  no 
constant  phase  angle  between  them;  but  this  angle  varies  peri- 
odically, so  that  at  some  intervals  of  time  the  two  waves  are 
nearly  in  phase,  and  at  others  they  are  nearly  in  opposition. 
The  familiar  method  of  synchronizing  two  alternators  by  means 
of  incandescent  lamps  is  based  on  this  phenomenon. 


CHAP.  IV]  SINE-WAVES  AND  VECTORS  35 

Prob.  1.  An  alternating  current  fluctuates  according  to  the  sine  law 
between  the  values  of  ±75  amp.,  making  6000  alternations  per  minute  - 
(3000  positive  and  as  many  negative  ones).  Draw  a  curve  of  instan- 
taneous values  of  this  current;  mark  on  the  axis  of  abscissae  the  time  t  in 
thousandths  of  a  second,  the  angles  u  in  degrees,  and  the  same  angles  in 
radians. 

Prob.  2.  What  is  the  frequency  of  the  current  in  the  preceding  probr 
lem,  in  cycles  per  second?  Ans.  50.  "^ 

Prob.  3.  Plot  on  the  same  curve  sheet  with  the  curve  obtained  in 
problem  1  the  sine-wave  of  a  current  the  frequency  of  which  is  three  times 
as  great,  and  the  amplitude,  52  amp.  The  curve  is  to  be  at  its  maximum 
when  the  first  curve  is  at  a  maximum. 

Prob.  4.  Supplement  the  preceding  curves  by  one,  the  frequency  of 
which  is  50  cycles  per  second,  the  amplitude  63  amp.,  and  which  reaches 
its  positive  maxima  at  the  same  instants  in  which  the  first  curve  passes 
through  zero.      Show  that  with  these  data  two  distinct  curves  can  be  ^ 
drawn. 

Prob.  6.  Draw  on  the  same  curve-sheet  with  the  preceding  curves  a 
sine-wave  representing  a  50-cycle  alternating  current,  the  amplitude  of 
which  is  120  amp.,  and  which  lags  by  30  degrees  with  respect  to  the 
current  in  problem  1. 

Prob.  6.  The  current  mentioned  in  problem  1  is  generated  by  a 
12-pole  alternator,  that  in  problem  3  by  a  14-pole  machine.  At  what 
speeds  must  these  machines  be  driven  in  order  to  give  the  required 
frequencies?  .  Ans.  500  and  1285  r.p.m. 

Prob.  7.  Express  the  currents  given  in  problems  1  to  5  by  equa- 
tions of  the  form  of  eq.  (36).  Ans.  i  =  75  sin  u;  i  =  —  52  sin  3  u;  i  = 
±63  cos  u;  i  =  120  sin  (u  —  30°). 

Prob.  8.  The  angle  u  in  the  answers  to  the  preceding  problem  is  ex- 
pressed in  degrees;  rewrite  the  equations  so  as  to  have  u  expressed  in 
radians,  and  in  fractions  of  a  cycle.  Also  represent  the  currents  as  func- 
tions of  the  time  t. 

Prob.  9.  Express  by  equations  similar  to  eq.  (41)  the  following  sinus- 
oidal voltages  of  frequency  /:  (a)  Amplitude  Em  volts,  (b)  Amplitude 
Em'  volts,  lagging  «  degrees  with  respect  to  the  first  curve,  (c)  Ampli- 
tude Em"  volts,  leading  the  second  curve  by  <t>  radians,  (d)  Amplitude 
Em'"  volts,  lagging  one  nth  of  a  cycle  with  respect  to  the  curve  (a). 

Prob.  10.  The  voltages  required  in  the  preceding  problem  are  induced        /' 
by  four  identical  alternators,  having  p  poles  each,  and  coupled  together. 
By  what  geometrical  angles  must  the  revolving  or  the  stationary  parts 
be  displaced  in  order  to  give  the  required  differences  in  phase? 

13.  Representation  of  a  Sine-wave  by  a  Vector.  It  is  clear 
from  the  foregoing  theory  and  problems  that  all  sine-wave  cur- 
rents or  voltages  are  different  from  one  another  in  three  respects 
only,  namely:  (1)  in  amplitude;  (2)  in  frequency;  and  (3)  in 
relative  phase  position.  In  most  practical  cases,  all  the  currents 


36  THE  ELECTRIC  CIRCUIT  [ART.  13 

and  voltages  entering  into  a  problem  are  of  the  same  frequency, 
so  that  they  differ  from  each  other  solely  in  their  amplitudes  and 
phase  positions.  In  such  cases  it  is  not  necessary  to  draw  sine- 
waves,  or  even  to  write  their  equations;  it  is  sufficient  to  indicate 
the  radii  /,„'  and  Im"  which  generate  these  curves  (Fig.  11),  in  their 
true  magnitudes  and  relative  positions.  The  rotating  radius,  at 
any  instant,  gives  by  its  vertical  projection  to  scale  the  magnitude 
of  the  alternating  current  or  voltage  at  that  instant. 

The  absolute  position  of  the  radii  is  immaterial,  because  they 
are  revolving  all  the  time.  It  is  their  relative  position  which  is 
permanent,  and  which  determines  the  relative  position  of  the 
sine-waves.  The  moment  from  which  time  is  counted  is  arbi- 
trary in  most  problems;  hence,  one  of  the  radii  can  be  drawn  in 
any  desired  position.  Then,  all  other  radii  in  the  same  problem 
are  determined  by  their  phase  displacement  with  respect  to  this 
"reference"  radius. 

It  must  be  clearly  understood  that  the  foregoing  representa- 
tion by  vectors  is  true  only  when  all  the  vectors  are  revolving  at 
the  same  speed,  that  is,  only  with  alternating  quantities  of  the 
same  frequency.  When  currents  and  voltages  of  different  fre- 
quencies enter  into  a  problem,  the  angle  between  the  vectors 
varies  all  the  time,  and  it  is  necessary  to  introduce  an  arbitrary 
zero  of  time  for  reference.  In  general,  the  graphical  method  of 
solution  is  unsuitable  for  such  problems. 

In  mathematics  and  physics,  a  quantity  which  has  not  only  a 
magnitude,  but  also  a  definite  direction  in  space  or  in  a  plane, 
is  called  a  vector.  Thus,  for  instance,  in  mechanics,  force  is  a 
vector  quantity,  while  volume  is  not.  The  radii  which  repre- 
sent sine-waves  have  both  magnitude  and  direction  in  a  plane. 
It  is  proper,  therefore,  to  call  them  vectors.  While  the  direction 
of  the  first  vector  is  usually  arbitrary,  once  it  is  selected,  the 
directions  of  all  the  other  radii  become  definite,  so  that  with 
this  limitation,  the  radii  in  alternating-current  problems  have 
definite  directions  and  may  be  called  vectors.  While  they  must 
be  imagined  as  revolving  when  generating  their  respective  sine- 
waves,  yet  they  revolve  as  a  system,  maintaining  their  relative 
positions  unchanged.  The  required  relations  always  depend 
upon  the  relative  positions  of  the  radii,  so  that  the  fact  that  they 
are  revolving  can  be  altogether  disregarded,  and  the  radii  consid- 
ered as  simple  stationary  vectors. 


CHAP.  IV]  SINE-WAVES  AND  VECTORS  37 

Prob.  1.  Draw  the  vectors  of  the  currents  in  problems  1,  4  and  5  of 
the  preceding  article  in  their  true  magnitudes  and  relative  positions. 

Prob.  2.  A  single-phase  alternator  has  a  terminal  voltage  of  which 
the  maximum  instantaneous  value  is  equal  to  16  kilovolts.  The  maxi- 
mum value  of  the  current  supplied  by  the  machine  is  325  amp.  The 
character  of  the  load  is  such  that  the  current  wave  lags  behind  the  voltage 
wave  by  an  angle  of  37  degrees.  Assuming  both  the  voltage  and  the 
current  to  vary  according  to  the  sine  Jaw,  represent  the  foregoing  condi- 
tions by  two  vectors.  y 

Prob.  3.  Draw  a  vector  diagram  showing  the  phase  (star)  voltages  and 
currents  of  a  25-cycle  three-phase  system  (Fig.  36),  the  amplitude  of 
each  voltage  being  7235  volts,  and  each  displaced  in  phase  by  120  degrees 
with  respect  to  the  other  two  voltages.  The  current  in  the  first  phase 
is  30  amp.,  and  lags  behind  the  corresponding  phase  voltage  by  £  of  a 
cycle.  The  current  in  the  second  phase  is  47  amp.,  and  leads  its  voltage 
by  18  degrees.  The  current  in  the  third  phase  is  72  amp.,  and  lags  behind 
the  corresponding  phase  voltage  by  0.004  of  a  second. 

Note:  In  the  foregoing  three  problems  the  student  is  supposed  to 
draw  the  vectors  equal  in  length  to  the  amplitudes  of  the  alternating 
waves.  In  practice,  it  is  customary  to  draw  vectors  equal  in  length  to 
the  effective  values  of  voltages  and  currents,  and  not  to  their  amplitudes. 
For  sine-waves  the  effective  value  is  equal  to  the  amplitude  divided  by 
\/2  (see  Chapter  5).  The  difference  is  not  important  for  our  present 
purposes.  The  use  of  effective  values  would  merely  change  the  arbi- 
trary scale  to  which  the  vectors  are  drawn. 

14.  Addition  and  Subtraction  of  Vectors.  There  are  many 
practical  problems  in  which  alternating  currents  or  voltages  have 
to  be  added,  or  subtracted  one  from  another.  For  instance, 
when  two  or  more  alternators  are  working  in  parallel,  the  total 
current  delivered  to  the  station  bus-bars  is  equal  to  the  sum  of 
the  currents  supplied  by  each  machine.  Or,  to  find  the  voltage 
at  the  receiving  end  of  a  transmission  line,  the  voltage  drop  in 
the  line  is  subtracted  from  the  generator  voltage.  When  the  com- 
ponent quantities  vary  according  to  the  sine  law  and  are  all  of  one 
frequency,  the  resultant  quantity  is  also  a  sine  curve  of  the  same 
frequency.  This  curve  may  be  found  (a)  graphically,  by  adding 
the  component  curves  point  by  point;  (b)  analytically,  by  adding 
their  equations;  or  (c)  by  adding  the  vectors  of  these  curves. 

It  must  first  be  proved  that  the  sum  of  two  sine- waves  of 
one  frequency  is  also  a  sine-wave  of  the  same  frequency.     Let 
the  two  currents  to  be  added  be  represented  by  the  equations 
i  =  Im  sin  (u  +  0 


i'  =  /„'  sin  (u  +  <j>'   " (42) 


38  THE   ELECTRIC  CIRCUIT  [ART.  14 

where  u  =  2  irft  is  the  variable  time  angle,  and  0  and  0'  are  two 
constant  angles  characterizing  the  relative  phase  positions  of  the 
two  waves  with  respect  to  some  reference  wave  Im"  sin  u.  The 
phase  displacement  between  i  and  i'  is  4>'  —  0.  Expanding  the 
foregoing  sines  of  the  sum  of  two  angles,  and  adding  the  two 
equations,  member  for  member,  we  obtain 

ieg  =  i  +  i'  =  (Im  cos  0  +  Im  cos  </>')  sin  u 

+  (Im  sin  0  +  Im'  sin  0')  cos  M,  .     .     .     .     (43) 

where  the  constant  coefficients  of  sin  u  and  cos  u  are  grouped 
together.  The  subscript  eq  stands  for  "equivalent."  This  ex- 
pression is  of  the  form  ieq  =  A  sin  u  +  B  cos  u,  where  A  and  B 
are  constants.  No  matter  what  values  A  and  B  may  have,  the 
right  hand  side  of  this  equation  is  reducible  to  the  form 

ieq  =  Ieqmsm(u  +  <i>eq) (44) 

Assuming  this  equation  to  be  true,  we  equate  the  right-hand  sides 
of  eqs.  (43)  and  (44),  and  expand  sin  (w  +  0eg).  Equating  the 
coefficients  of  sin  u  and  cos  u,  we  get 

Iegm  cos  0ca  =  Im  cos  0  +  /„/  cos  0';  1 
leqm  sin  0eg  —  Im  sin  0  +  Im'  sin  0'.  ) 

These  are  two  simultaneous  equations  with  Ieqm  and  0e9  as  the 
unknown  quantities.  Squaring  and  adding  these  equations,  we 
obtain 

/e7m2  =  (/«  sin  0  +  Im'  sin  0')2  +  (/„  cos  0  +  7TO'  cos  0')2-     (46) 
Dividing  the  second  equation  by  the  first  gives 

tan  0C,  =  (Im  sin  0  +  /„/  sin  0')/(/«  cos  0  +  7m'  cos  0')-     (47) 

No  matter  what  values  Im,  Im',  0  and  0'  may  have,  the  values 
of  Ieqm  and  0e,  determined  from  these  equations  are  real.  In 
other  words,  it  is  always  possible  to  represent  eq.  (43)  in  the 
form  of  eq.  (44).  This  proves  the  proposition,  because  we  see 
from  eq.  (44)  that  ieq  is  a  sine-wave  having  the  same  u  =  2  irft  for 
the  variable  angle,  hence  the  same  frequency  as  the  component 
waves.  The  amplitude  and  the  phase  position  of  this  resultant 
wave  are  determined  by  eqs.  (46)  and  (47) . 

When  two  currents  or  voltages  are  represented  by  vectors, 
their  sum  or  difference  is  also  a  vector,  because,  as  proved  before, 


CHAP.  IV.] 


SINE-WAVES  AND  VECTORS 


39 


it  is  also  a  sine-wave  of  the  same  frequency.  The  problem 
is  to  find  the  vector  of  the  resultant  wave,  knowing  the  vectors 
of  the  component  waves  in  their  magnitudes  and  positions.  Any 
ordinate  of  the  resultant  wave  must  be  equal  to  the  sum  of  the 
corresponding  ordinates  of  the  component  waves.  Hence,  the 
vector  of  the  resultant  wave  must  satisfy  the  condition  that  its 
projection  upon  the  F-axis  (Fig.  12)  shall  be  equal  to  the  sum  of 
the  projections  of  the  component  vectors  on  the  same  axis.  This 
condition  must  be  fulfilled  at  all  instants  of  time,  that  is,  during 
the  rotation  of  the  three  vectors.  To  satisfy  this  requirement 
the  resultant  vector  must  be  the  diagonal  of  a  parallelogram  of 
which  the  other  two  vectors  are  the  adjacent  sides. 


FIG.  12.    Addition  of  vectors. 

Let  OA  and  OB  be  the  given  vectors  to  be  added  together. 
From  the  end  B  of  the  vector  OB  draw  a  line  BC  equal  and  paral- 
lel to  OA.  Connecting  0  and  C  gives  the  resultant  vector  OC, 
in  magnitude  and  position.  It  will  be  seen  from  the  figure  that 
the  projection  of  OC  upon  the  F-axis  is  equal  to  the  sum  of  the 
projections  of  OB  and  BC  upon  the  same  axis.  But  BC  is  equal 
and  parallel  to  OA,  so  that  the  projection  of  OC  on  the  vertical 
axis  is  equal  to  the  sum  of  the  projections  of  the  given  vectors 
on  the  same  axis.  This  construction  holds  true  for  any  instant 
whatever.  By  drawing  AC,  the  parallelogram  OBCA  is  com- 
pleted, so  that  the  construction  is  identical  with  that  for  finding 
the  resultant  of  two  mechanical  forces.  However,  in  practical 
applications  it  is  not  necessary  to  complete  the  parallelogram, 
because  the  resultant  is  perfectly  determined  by  the  triangle 
OBC.  The  resultant  of  two  vectors  obtained  in  this  way  is  called 
their  geometric  sum. 


40 


THE  ELECTRIC  CIRCUIT 


[ART.  14 


If  the  triangle  were  not  closed,  the  condition  of  equality  with 
the  sum  of  the  projections  of  the  given  vectors  might  be  satisfied 
for  one  particular  instant  of  the  cycle,  but  would  not  be  satisfied 
for  other  instants.  Thus,  for  instance,  assuming  the  line  OC' 
to  be  the  resultant  vector,  we  see  that  for  the  instant  shown  in 
the  sketch  the  projection  of  OC'  upon  the  axis  OF  is  equal  to  the 
sum  of  the  projections  of  OB  and  BC  upon  the  same  axis;  but  the 
condition  is  not  fulfilled  when  the  vectors  rotate. 

The  rule  for  subtraction  of  vectors  follows  immediately  from 
the  preceding  rule,  because  to  subtract  a  vector  means  to  add  a 
vector  with  the  opposite  sign.  Thus,  let  it  be 
required  to  subtract  the  vector  OA  from  OB 
(Fig.  13);  this  may  mean,  for  instance,  the 
subtraction  of  the  voltage  wave  represented 
by  OA  from  that  represented  by  OB.  From 
the  end  B  of  OB  draw  vector  BC  equal  and 
opposite  to  OA.  The  resultant,  OC,  represents 
the  difference  of  the  two  given  vectors,  in 
direction  and  magnitude,  and  thus  determines 
the  sine-wave  of  the  resultant  voltage.  If  it 
were  required  to  subtract.  OB  from  OA,  it 
would  be  necessary  to  draw  AC'  equal  and 
opposite  to  OB,  thus  obtaining  the  resultant 
OC',  equal  and  opposite  to  the  former  resultant 
OC.  This  is  in  accord  with  the  general  alge- 
FIG.  13.  Subtraction  braic  rule  that  A  -  B  =  -(B  -A}. 

The  preceding  results  with  regard  to  the 
addition  and  subtraction  of  vectors  are  summed  up  in  the  follow- 
ing rule :  Relations  which  are  true  algebraically  for  instantaneous 
values  of  sinusoidal  currents  and  voltages,  hold  true  geometrically 
for  the  vectors  of  these  quantities.  It  is  customary  to  provide 
vectors  of  currents  with  triangular  arrows,  as  in  Fig.  12;  vectors 
of  voltages  are  usually  distinguished  by  pointed  arrows,  as  in 
Fig.  13.  This  distinction  enables  one  to  see  directly  from  the 
diagram  whether  a  vector  represents  a  current  or  a  voltage,  with- 
out reference  to  the  text. 


Prob.  1.  The  currents  generated  by  two  alternators  in  parallel  are 
75  and  120  amp.  respectively,  the  second  current  lagging  behind  the  first 
by  30  degrees.  Determine  the  magnitude  and  the  relative  phase  position 


CHAP.  IV]  SINE-WAVES  AND  VECTORS  41 

of  the  resultant  line  current  by  three  methods:    (a)  point  by  point;    i/ 
(b)  analytically;    (c)  by  means  of  vectors. 

Ans.     188.8  amp.,  lagging  by  18°  32'  behind  the  first  current. 

Prob.  2.  Solve  the  preceding  problem  without  the  use  of  eqs.  (46)    ^/ 
and  (47),  simply  by  means  of  the  theorem  proved  above,  that  the  sum 
or  the  difference  of  two  sine- waves  is  also  a  sine- wave.    Solution: 

leg  sin  (u  +  <t>eq)  =  75  sin  u  +  120  sin  (u  -  30°). 

This  equation  is  true  for  any  instant,  or  for  any  value  of  u.  It  contains 
two  unknown  quantities,  the  amplitude  and  the  phase  position  of  the 
resultant  curve.  It  is  necessary,  therefore,  to  apply  this  equation  to  two 
particular  moments  of  time,  in  order  to  obtain  two  equations  with  two 
unknown  quantities.  It  is  most  convenient  in  this  particular  case  to 
choose  u  =  7I-/2  and  u  =  0.  Substituting  these  values,  two  equations 
with  two  unknown  quantities  are  obtained.  This  method  is  preferable 
in  the  solution  of  practical  problems,  because  it  is  not  necessary  to 
remember  eqs.  (46)  and  (47),  and  also  because  the  two  values  of  u  can 
be  selected  so  as  to  give  the  simplest  equations. 

Prob.  3.  Two  alternators,  with  the  same  number  of  poles,  are  coupled 
together  so  as  to  give  voltages  differing  in  phase  by  27  degrees,  the  voltage 
of  the  second  machine  leading  that  of  the  first.  The  first  alternator 
generates  a  voltage  the  amplitude  of  which  is  2300  volts,  the  second 
1800  volts.  The  two  machines  are  connected  electrically  in  series.  Find 
graphically  the  vector  of  the  resultant  voltage  in  its  magnitude  and  phase 
position.  Find  also  the  vector  of  the  resultant  voltage  when  the  termi- 
nals of  one  of  the  machines  are  reversed.  Ans.  (1)  3988  volts,  lead- 
ing the  first  by  11°  49';  (2)  1074  volts,  lagging  behind  the  first  by  49°  32'. 

Prob.  4.  An  alternator,  the  terminal  voltage  of  which  is  6600,  supplies 
its  load  through  a  transmission  line.  The  conditions  are  such  that  the 
current  lags  behind  the  generator  voltage  by  an  angle  of  35  degrees. 
The  voltage  drop  in  the  line  is  540  volts,  leading  the  current  ia -phase  by 
an  angle  of  67  degrees.  Find  the  receiver  voltage  by  subtracting  the 
voltage  drop  in  the  line  from  the  generator  voltage  (geometrically) ;  also 
determine  the  phase  displacement  between  the  receiver  voltage  and  the 
current.  Ans.  6149  volts;  32°  20'. 

15.  Non-sinusoidal  Currents  and  Voltages.  When  a  current 
or  voltage  wave  differs  considerably  from  the  pure,  sine  form, 
it  is  often  convenient  to  represent  it  as  the  result  of  a  superposi- 
tion of  sine- waves  of  different  frequencies  (Fig.  14).  No  mat- 
ter how  complicated  a  periodic  wave  may  be,  it  can  always  be 
so  represented  with  sufficient  accuracy,  by  properly  selecting  the 
amplitudes  and  the  phase  relations  of  the  component  sine-waves, 
o'r  harmonics,  as  they  are  called.  Theoretically,  an  infinite  num- 
ber of  sine-waves  is  necessary  in  order  to  represent  any  given 
irregular  wave  exactly.  In  practice,  however,  a  limited  number 
of  harmonics  is  sufficient. 


42 


THE  ELECTRIC  CIRCUIT 


[ART.  15 


CHAP.  IV]  SINE-WAVES  AND  VECTORS  43 

If  the  frequency  of  the  given  irregular  wave  is/,  the  frequencies 
of  the  harmonics  are  /,  2  /,  3  /,  and  so  on.  When,  however,  the 
given  wave  is  symmetrical,  that  is,  when  the  part  above  the  axis 
of  abscissae  is  identical  with  that  below,  all  even  harmonics  (2/, 
4/,  etc.)  drop  out,  and  the  wave  consists  only  of  the  fundamental 
wave  of  frequency  /,  and  the  odd  harmonics  (3/,  5/,  etc.).  The 
student  can  easily  convince  himself  of  the  truth  of  this  statement 
by  taking  a  fundamental  sine-wave  and  adding  to  it  a  second 
harmonic  and  a  third  harmonic.  In  the  first  case  the  resultant 
wave  will  be  unsymmetrical;  in  the  second,  symmetrical.  Nearly 
all  of  the  waves  encountered  in  practice  are  symmetrical. 

Let  the  fundamental  wave  be  represented  by  the  equation 
7/1  =  Ci  sin  (u  —  ai),  the  third  harmonic  by  the  equation  y$  = 
Cs  sin  3  (u  —  as),  etc.  The  meaning  of  Ci,  €3,  etc.,  and  of  ai, 
a3,  etc.,  is  clear  from  Fig.  14;  0  is  an  arbitrary  origin  from  which 
the  angles  are  measured.  The  ordinates  of  the  given  composite 
symmetrical  wave  are  represented  by  the  equation 

y  =  Ci  sin  (u  —  ai)  +  C3  sin  3  (u  —  a3)  +  C5  sin  5  (u  —  «5) 

+  etc (48) 

This  expression  is  known  as  the  Fourier  series,  and  is  of  great 
importance  in  mathematical  physics. 

In  practice,  the  problem  which  presents  itself  is  usually  that  of 
analysis;  that  is  to  say,  it  is  often  required  to  analyze  or  resolve 
a  given  irregular  wave  into  its  harmonics.  In  other  words,  know- 
ing y,  one  is  asked  to  determine  the  values  of  C  and  a  for  one  or 
more  harmonics.  This  is  a  purely  mathematical  problem,  and  is 
not  treated  here,  because  the  solution  will  be  found  in  numerous 
textbooks,  handbooks  and  magazine  articles.1  There  are  also 
mechanical  wave  analyzers  on  the  market,  by  means  of  which 
any  desired  harmonic  may  be  separated  by  tracing  the  given 
curve  with  a  stylus,  in  a  manner  similar  to  the  way  in  which  a 
planimeter  is  used. 

It  is  of  importance  for  an  electrical  engineer  to  train  his  eye 
in  the  discernment  of  prominent  harmonics,  without  mathematical 
analysis.  This  training  is  afforded  by  exercises  in  wave  synthesis, 
that  is,  in  combining  various  assumed  harmonics  into  irregular 
waves. 

1  See,  for  instance,  the  author's  Experimental  Electrical  Engineering,  Vol.  2, 
p.  222. 


44  THE  ELECTRIC  CIRCUIT  [ART.  15 

Take  first  a  fundamental  sine-wave  and  a  third  harmonic  of 
a  reasonable  magnitude,  say  between  15  and  30  per  cent  of  the 
fundamental.  Combine  these  waves  into  one,  with  different  rela- 
tive phase  positions  of  the  fundamental  and  the  third  harmonic. 
In  this  way  a  flat  wave,  a  peaked  wave  and  a  one-sided  "humped" 
wave  will  be  obtained.  Then  change  the  magnitude  of  the  third 
harmonic  and  construct  similar  waves,  in  order  to  see  the  influ- 
ence of  this  factor.  After  that,  plot  similar  curves  for  the  funda- 
mental wave  with  a  fifth  harmonic,  a  seventh  harmonic,  and  so 
on.  Finally,  combine  the  fundamental  wave  with  the  third  and 
the  fifth  harmonics  simultaneously,  and  so  on.  After  some  prac- 
tice, the  eye  will  easily  discern  prominent  harmonics  in  a  given 
irregular  wave.  Numerous  oscillograms  of  irregular  waves  will 
be  found  in  many  current  periodicals  and  in  the  transactions  of 
the  various  electrical  societies.  Read  in  this  connection  Art.  30 
of  the  Magnetic  Circuit. 

Prob.  1.  Draw  two  or  three  sets  of  curves  suggested  in  the  preceding 
paragraph;  each  set  must  comprise  about  six  curves  and  each  curve 
must  have  a  harmonic  with  a  different  phase  position. 

Prob.  2.  Devise  a  simple  apparatus  by  means  of  which  harmonics 
can  be  combined  mechanically,  and  the  resultant  waves  observed,  with- 
out actually  plotting  curves  point  by  point. 

Prob.  3.  Analyze  a  given  irregular  wave  into  its  harmonics,  using  the 
method  given  in  the  reference  above,  or  any  other  method  found  in  the 
literature  on  the  subject. 


CHAPTER  V 
POWER  IN  ALTERNATING-CURRENT  CIRCUITS 

V     16.  Power  when  Current  and  Voltage  are  in  Phase.     Let 

a  resistance  r  be  connected  across  the  terminals  of  an  alternator, 
the  voltage  at  the  terminals  varying  according  to  the  sine  law. 
The  current  through  the  resistance  also  varies  according  to  the 
sine  law,  because  Ohm's  law  holds  true  for  any  moment  of  time, 
so  that  the  curve  of  the  current  is  in  phase  with  that  of  the  voltage. 
If  the  equation  of  the  voltage  wave  is  e  =  Em  sin  u,  the  equation 
of  the  current  is  i  =  (Em/r)  •  sin  u.  Graphically,  the  current  and 
the  voltage  are  represented  by  two  vectors  of  different  lengths,  but 
in  the  same  direction  —  for  instance,  like  OC  and  OD  in  Fig.  13. 

Divide  the  time  T  of  one  cycle  into  a  large  number  of  small 
intervals  A£.  Then  the  amount  of  energy  delivered  to  the  resist- 
ance r  and  converted  into  joulean  heat  during  one  of  such  inter- 
vals varies  with  the  time  position  of  the  interval  in  the  cycle, 
in  other  words,  with  the  instantaneous  values  of  the  voltage  and 
the  current.  This  energy  is  practically  equal  to  zero  when  the 
current  and  the  voltage  have  values  near  zero,  and  it  reaches  a 
maximum  with  them.  However,  the  dissipated  energy,  being  in 
the  nature  of  a  frictional  loss,  never  becomes  negative,  because 
whether  the  current  flows  in  one  direction,  or  in  the  other,  the 
heat  liberated,  e  •  i  •  AZ  =  i*r  •  A£,  is  always  positive. 

Since  the  voltage  and  the  current  vary  with  the  time,  the  rate 
of  liberation  of  energy,  or  the  instantaneous  power,  is  also  variable. 
The  expression  P  =  ei  =  izr  represents  the  instantaneous  power 
as  with  direct  current.  If  e  and  i  remained  constant  for  one 
second,  the  energy  liberated  would  be  equal  to  i2r.  As  a  matter 
of  fact,  e  and  i  may  be  considered  constant  only  during  the  infini- 
tesimal element  of  time  dt,  so  that  the  energy  liberated  during  the 
time  dt  is  i2r  •  dt.  Nevertheless,  it  is  proper  to  say  that  at  the  in- 
stant under  consideration  the  energy  is  liberated  at  a  rate  equal  to 
izr  per  second,  because  (izr  •  dt}/dt  =  i2r.  This  is  analogous  to  the 

45 


46  THE  ELECTRIC  CIRCUIT  [ART.  16 

way  in  which  we  speak  of  the  instantaneous  speed  of  a  body  during 
a  period  of  acceleration  or  retardation.  The  speed  varies  from 
instant  to  instant,  so  that  to  say  that  the  speed  is  v  at  a  certain 
instant  merely  means  that,  if  the  body  continued  to  move  at  this 
velocity  for  one  second,  it  would  cover  a  space  equal  to  v.  In  the 
same  sense,  the  instantaneous  power  indicates  the  amount  of 
energy  which  would  be  developed  per  second,  if  the  current  and 
the  voltage  suddenly  became  constant. 

The  total  energy  liberated  in  the  form  of  heat  during  one 
complete  cycle  is 


nT  r*T 

W  =  I     izr-dt  =  r  I     i2-dt. 
Jo  Jo 


(49) 


When  no  local  e.m.fs.  are  present,  the  same  energy  is  repre- 
sented by  the  expression 

W  =  fT  ei-dt  .......     (50) 

Jo 

When  there  are  local  e.m.fs.  in  the  part  of  the  circuit  under 
consideration,  the  total  energy  communicated  to  it  during  an 
interval  of  time  is  different  from  that  dissipated  as  heat  (Art.  4). 
According  to  eq.  (19),  we  have 


W  = 


reii-dt  =  r  f 

Jo  Jo 


Suppose,  for  example,  that  e\  is  the  counter-e.m.f.  of  a  motor 
in  the  circuit,  and  therefore  nearly  in  phase  opposition  to  et.  Then 
the  i'V  loss  on  the  right-hand  side  of  the  equation  is  the  difference 
between  the  energy  supplied  to  the  circuit  and  that  converted  into 
mechanical  work  in  the  motor. 

The  foregoing  equations  are  true  whether  the  current  and  the 
voltage  vary  according  to  the  sine  law  or  not.  If  they  are  sinus- 
oidal, the  integration  can  be  easily  performed,  and  the  energy  per 
cycle  evaluated  by  the  following  method.  Let  the  current  be 
represented  as  before  by  i  =  Im  sin  u.  Substituting  this  value 
into  eq.  (49),  we  have 

W  =  Imzr  fTsmzudt  ......     (52) 

This  expression  is  easily  integrated  by  using  the  substitution 
sin'u  =  £(1  —  cos2u).  Or  it  may  be  evaluated  by  observing 
that  its  value  remains  the  same  if  a  cosine  is  substituted  for  the 


CHAP.  V]    POWER  IN  ALTERNATING-CURRENT  CIRCUITS          47 

sine.  This  is  because  the  limits  of  integration  are  u  =  0  and 
u  =  2  TT,  and  in  summing  up  sines  or  cosines  through  2  TT  we  take 
the  same  quantities,  only  in  a  different  order.  Hence  we  may 
write 

W  =  In?r  fTcos?udt.  (52a) 

Jo 

Adding  the  two  expressions  term  by  term,  and  remembering  that 
sin2  w  -j-  cos2  u  =  1,  we  get 

2W  =  Im*r  CTdt  =  Imzr.T, 
Jo 

or  the  energy  converted  into  heat  during  one  cycle  is 

W  =  lIm*r-T (53) 

When  there  are  no  local  e.m.fs.  and  the  current  is  in  phase 
with  the  voltage,  we  have  Em  =  Imr,  so  that  from  eq.  (53),  and  by 
analogy  to  eq.  (18),  we  have 

W  =  %ImEm-T (54) 

and 

W  =  %(Em*/r).T (55) 

The  student  must  clearly  understand  that  the  phase  relation 
between  the  current  and  the  voltage  is  of  no  consequence  in  eq. 
(53),  while  eqs.  (54)  and  (55)  hold  true  only  when  the  current  is 
in  phase  with  the  voltage.  Or  else,  Em  in  these  latter  expressions 
may  be  said  to  refer  to  that  component  of  the  total  terminal 
voltage  which  is  used  up  in  Ir  drop. 

Prob.  1.   A  sine-wave  alternating  current,  which  fluctuates  between 
±75  amp.,  flows  through  a  resistance  of  10  ohms.     Plot  curves  of  instan- 
taneous values  of  the  voltage  and  power;  the  frequency  is  50  cy./sec. 
Ans.    Em  =  750  volts;  max.  power  =  56.25  kw. 

Prob.  2.   Determine  the  total  energy  liberated  per  cycle  in  the  pre-    \ 
ceding  problem,  by  integrating  graphically  the  curve  of  power. 

Ans.    562.5  joules  (watt-seconds). 

Prob.  3.  Prove  analytically  that  the  curve  of  power  obtained  in 
problem  1  is  a  sine-wave  of  double  frequency,  tangent  to  the  axis  of 
time.  Proof:  The  equation  of  the  curve  is  P  =  7mV  •  sin2  u.  But  from 
trigonometry  . 

cos  2  u  =  cos2  u  —  sin2  u  =  1  —  2  sin2  M. 

Substituting  the  value  of  sin2  u  from  this  equation  into  the  expression  for 
P,  we  get 

P  =  I  Infr  -  i  Ir,?T  •  COS  2  U. 

The  first  term  is  constant,  while  the  second  represents  a  sine-wave  of 


48  THE  ELECTRIC  CIRCUIT  [ART.  17 

double  frequency,  because  2  u  =  2  IT  (2  f)t.  The  first  term  is  never 
smaller  than  the  second,  so  that  P  is  always  positive,  and  the  whole  curve 
lies  above  the  axis  of  abscissae.  The  second  term  becomes  equal  to  the 
first  and  P  =  0,  only  when  2  u  is  a  multiple  of  2  IT.  At  these  points  the 
curve  is  tangent  to  the  axis  of  abscissa;. 

Prob.  4.   Deduce  eq.  (53)  directly  from  (52),  expressing  sin  u  in  terms 
of  the  cosine  of  the  double  angle,  as  in  the  preceding  problem.     Hint:     . 
From  eq.  (37),  dt  =  (T/2  TT)  du,  and  the  limits  of  integration  are  u  =  0  and 


17.  The  Effective  Values  of  Current  and  Voltage.  In  prac- 
tice, it  is  the  average  rate  of  delivery  or  dissipation  of  energy 
that  is  of  interest,  or,  in  other  words,  the  average  value  of  the 
variable  instantaneous  power.  This  is  analogous  to  using  in 
calculations  the  average  speed  of  a  machine,  when  the  actual 
speed  varies  within  certain  limits.  This  average  power  is  found 
by  dividing  the  total  energy  developed  during  one  cycle  by  the 
period  T  of  the  cycle.  When  the  current  varies  according  to 
the  sine  law,  the  total  energy  per  cycle  converted  into  heat  is 
expressed  by  eq.  (53).  Dividing  both  sides  by  T,  we  find  that 
the  average  power 

Pave   =   \  In?r  ........        (56) 

It  is  convenient  to  use  in  eq.  (56)  a  new  value  of  the  current, 

7  =  7m/v/2  =  0.707  Im,      .....     (57) 

instead  of  7m,  because  then  the  expression  for  the  average  power 
becomes  identical  with  that  in  a  direct-current  circuit,  namely, 

Pa*e  =  I2r  ........     (58) 

Analogously,  if  we  define 

E  =  Em/V2  =  0.707  Em,    .....     (59) 

eqs.  (54)  and  (55)  become 

Pave  =  E'I      .......     (60) 

and 

.......     (61) 


which  are  perfectly  similar  to  the  corresponding  expressions  in  a 
direct-current  circuit. 

E  and  7,  as  defined  above,  are  called  the  effective  values  of  the 
alternating  voltage  and  current  respectively.  We  may  say  that 
by  definition  the  effective  value  of  an  alternating  (or  variable) 
current  is  equal  to  such  a  constant  current  which,  when  flowing 


CHAP.  V]    POWER  IN  ALTERNATING-CURRENT  CIRCUITS          49 

through  a  resistance,  dissipates  the  same  average  power  as  the 
actual  variable  current. 

This  definition  of  an  effective  value  applies  to  variable  cur- 
rents of  any  form.  It  is  used,  for  instance,  in  determining  the 
temperature  rise  of  electric  railway  motors.  During  the  run  of  a 
car  the  current  fluctuates  within  wide  limits,  but  the  heating  of 
the  motor  windings  is  nearly  the  same  as  would  occur  with  a 
certain  constant  current,  which  is  called  the  effective  value  of 
the  actual  variable  current.  The  condition  for  the  same  average 
i2r  loss  is 


Pr-T 


=  r  C 


where  T  is  the  interval  of  time  for  which  it  is  desired  to  obtain 
the  effective  value.     Hence 


C 

Jo 


(62) 


This  equation  expresses  in  mathematical  language  that  72  is  the 
average  value  of  i2,  over  the  period  of  time  T.  Taking  the  square 
root  of  both  sides  of  this  equation,  we  can  also  define  the  effective 
value  /  as  the  square  root  of  the  mean  square  of  the  instantaneous 
values.  This  definition  is  true  for  any  form  of  alternating  or 
variable  current.  The  effective  voltage  is  defined  by  a  similar 
expression,  so  that  more  generally 


f 

Jo 


(63) 


where  y  denotes  an  instantaneous  value  of  current  or  voltage. 
Alternating-current  ammeters  and  voltmeters  are  always  cali- 
brated so  as  to  indicate  the  effective  values  of  current  and  voltage. 
When  an  irregular  wave  of  current  or  voltage  is  given  graphi- 
cally, its  effective  value  is  found  by  taking  a  sufficient  number 
of  equidistant  ordinates  (Fig.  15)  and  replacing  the  integration 
in  eq.  (63)  by  a  summation.  Let  the  half-wave  be  divided  into 
k  equal  parts,  where  k  is  an  even  number,  and  let  yo,  yi,  .  .  .  yk 
be  the  corresponding  ordinates.  Then,  according  to  Simpson's 
Rule, 


(63a) 


50 


THE  ELECTRIC  CIRCUIT 


[ART.  17 


The  larger  the  number  of  ordinates,  the  more  accurate  is  the  value 
of  y,ff  determined  by  this  method.  The  value  of  ye/f2  may  also 
be  found  by  plotting  a  curve  of  y2  against  u,  as  shown  in  Fig.  15, 
and  determining  its  mean  ordinate  by  means  of  a  planimeter. 

When  a  large  number  of  effective  values  must  be  determined, 
—  for  instance,  from  the  records  obtained  by  a  graphic  ammeter 
during  several  runs  of  an  electric  train,  —  the  squaring  of  ordi- 
nates becomes  a  tedious  process.  Some  practical  methods,  by 
means  of  which  the  necessity  for  squaring  ordinates  is  eliminated, 
are  described  in  the  next  article. 


FIG.  15.    The  effective  and  the  average  ordinates  of  an  irregular  half-v 


The  effective  value  of  a  current  or  a  voltage  is  also  called  the 
quadratic  mean,  to  distinguish  it  from  the  arithmetical  mean  value 
defined  by  the  familiar  equation 


-i 


or  for  a  periodic  curve 


(64) 


(65) 


It  will  be  noted  that  the  upper  limit  of  integration  is  TT  and 
not  2  TT.  It  is  evident  that  for  a  symmetrical  wave  the  average 
ordinate  over  a  whole  cycle  is  equal  to  zero.  The  average  value, 
therefore,  always  refers  to  a  half-wave. 


CHAP.  V]    POWER  IN  ALTERNATING-CURRENT  CIRCUITS          51 

W  =  T>)».J&~  "*     ~    $ 

For  the  sine-wave 

sin  u  du  =  (2/ir)ym,    - 

or 

yave/ym  =  2/7T  =  0.637 (66) 

The  ratio  of  the  effective  to  the  mean  ordinate  is  called  the 
form  factor,  because  it  gives  an  idea  of  the  degree  to  which  the 
curve  is  flat  or  peaked  as  compared  to  the  sine-wave.  For  a 
sine-wave  the  form  factor  is 

(7/m/V2)/(2i/m/7r)  =  l.ll (67) 

For  a  perfectly  flat-topped  or  rectangular  wave,  the  maximum 
value,  the  effective  value  and  the  average  value  are  all  the  same, 
so  that  the  form  factor  is  equal  to  unity.  For  very  peaked  waves, 
the  influence  of  the  high  middle  ordinates  is  more  prominent  in 
the  quadratic  mean,  so  that  the  effective  is  considerably  higher 
than  the  mean  value,  and  the  form  factor  is  larger  than  1.11. 

Another  ratio  which  helps  in  judging  about  the  shape  of  a 
curve  is  the  so-called  amplitude  factor,  or  the  ratio  of  the  maximum 
ordinate  to  the  effective  value.  The  author  is  not  aware  that 
either  the  form  factor  or  the  amplitude  factor  is  used  to  any  con- 
siderable extent  in  practice. 

Prob.  1.  An  electric  heater  was  tested  for  power  consumption  on  an 
alternating-current  circuit,  by  having  an  ammeter  in  series  with  it,  and 
a  voltmeter  across  its  terminals.  Both  instruments  were  calibrated  to 
indicate  effective  values.  The  readings  were  110  volts  and  5.7  amp. 
Assuming  the  current  and  the  voltage  to  have  been  in  phase,  which  is 
nearly  the  case,  what  was  the  average  power  consumption  of  the  heater, 
and  what  was  its  resistance?  Determine  also  the  maximum  instan- 
taneous values  of  the  current  and  the  voltage,  under  the  supposition  of 
the  sine  law.  Ans.  627  watts;  19.3  ohms;  155.56  volts;  8.06  amp. 


FIG.  16.    A  stepped  curve  of  current  or  voltage. 

Prob.  2.  Determine  the  average  value,  the  effective  value,  the  form 
factor  and  the  amplitude  factor  of  the  curve  shown  in  Fig.  16. 

Ans.    0.75  ym;  0.791  ym;  1.055;  1.264. 


52 


THE  ELECTRIC  CIRCUIT 


[ART.  18 


Prob.  3.  Check  sonic  of  the  values  of  the  form  factor  and  the  ampli- 
tude factor  given  in  the  table  in  the  Standard  Handbook  (see  Index  under 
"  form  factor").  This  will  afford  practice  in  calculating  effective  values 
of  curves  when  they  are  given  by  analytic  equations  of  the  form  y  =  f(t), 
using  eq.  (63). 

Prob.  4.  Plot  an  irregular  wave,  taken  from  an  available  oscillograph 
record,  and  calculate  its  average  and  effective  values  by  the  point-by- 
point  method,  or  by  using  a  planimeter. 

v 

18.  Some  Special  Methods  for  Calculating  the  Effective 
Value  of  an  Irregular  Curve.  As  is  mentioned  in  the  preceding 
article,  squaring  a  large  number  of  ordinates  in  order  to  find 
the  effective  value  of  a  curve  is  a  tedious  process,  and  methods 
are  available  which  sometimes  lead  to  the  end  more  quickly.  It 
must  be  admitted,  however,  that  for  one  who  has  to  do  this 
work  only  occasionally,  the  plain  point-by-point  method  described 
above  is  probably  the  quickest  and  the  most  reliable. 


FIG.  17.    An  irregular  curve  (Fig.  15)  and  the  equivalent  sine-wave, 
plotted  in  polar  coordinates. 

(a)  Fleming's  Method.  The  given  curve  (Fig.  15)  is  replotted 
in  polar  coordinates  (Fig.  17),  so  that  equal  polar  angles  O/fc  cor- 
respond to  equal  distances  ir/k  upon  the  axis  of  absciss®.  The 
ratio  between  an  abscissa  u  in  Fig.  15  and  the  corresponding  polar 
angle  in  Fig.  17  is  of  no  consequence;  in  other  words,  it  makes 
no  difference  what  total  central  angle  fl  corresponds  to  the  total 
distance  TT  in  Fig.  15.  The  area  of  an  infinitesimal  triangle  sub- 
tended by  a  polar  angle  du  is  \  i/2do>,  because  y  is  the  base  of  the 


CHAP.  V]     POWER  IN  ALTERNATING-CURRENT  CIRCUITS  53 

triangle,  and  y  dco  is  its  altitude.  Thus,  the  total  area  of  the  curve 
in  Fig.  17  is 

S-|/V<fe  ........     (68) 

But,  by  the  defining  equation  (63),  the  effective  value,  or  the  quad- 
ratic mean  ordinate,  of  the  same  curve  in  Fig.  15  is  found  from 
the  expression 

(69) 


because  co  is  proportional  to  t.  Comparing  the  preceding  two 
equations,  we  find  that 

1^  =  25/0  ........     (70) 

Since  S  is  easily  evaluated,  for  instance,  by  means  of  a  planimeter, 
the  effective  value  is  calculated  from  eq.  (70)  without  squaring 
the  ordinates,  but  simply  by  replotting  the  given  curve  in  polar 
coordinates.1 

When  the  given  curve  is  a  pure  sine-wave,  the  corresponding 
curve  in  polar  coordinates  is  a  circle,  provided  that  the  angle  fi 
is  selected  equal  to  IT.  The  student  can  easily  prove  this  for  him- 
self, either  graphically  or  analytically.  Let  ym  be  the  maximum 
ordinate  of  the  sine-curve;  then  the  area  of  the  circle  is  S  =  j  irymz, 
and  from  eq.  (70)  we  find  yeff  =  ym/\/2.  This  is  the  same  value 
as  found  before  by  a  different  method. 

When  the  given  curve  is  not  much  different  from  a  pure  sine- 
wave,  the  corresponding  polar  curve  approaches  a  circle  in  form 
(always  provided  that  O  =  T).  In  such  cases  it  is  possible  to 
determine  the  area  of  the  polar  curve  without  a  planimeter,  by 
drawing  a  circle  of  equal  area  as  judged  by  the  eye  (Fig.  17). 
The  effective  value  is  then  the  same  for  the  given  curve  and  for 
the  sine-wave  corresponding  to  this  circle,  and  is  equal  to  the 
diameter  of  the  circle  divided  by  V2.  Such  a  sine-wave  is  called 
the  equivalent  sine-wave.  It  is  often  convenient  in  dealing  with 
irregular  current  and  voltage  waves  to  replace  them  by  equivalent 
sine-waves,  so  as  to  be  able  to  apply  an  analytical  solution,  or 
to  construct  vectors. 

1  For  a  more  detailed  treatment  and  numerous  practical  applications, 
see  C.  O.  Mailloux,  "  Me'thode  de  Determination  du  Courant  Constant  Pro- 
duisant  le  meme  Echauffement  qu'un  Courant  Variable,"  in  the  Transac- 
tions of  the  International  Congress  of  Applications  of  Electricity,  Turin,  1911. 


54  THE  ELECTRIC  CIRCUIT  [ART.  18 

(b)  The  Effective  Value  in  Terms  of  Harmonics.  When  an 
irregular  wave  is  given  in  the  form  of  a  Fourier  series,  eq.  (48), 
the  effective  value  can  be  expressed  through  the  amplitudes  of 
the  harmonics.  In  order  to  use  the  expression  for  y  in  the  funda- 
mental formula  (63),  we  have  to  square  the  Fourier  expansion. 
This  gives  terms  of  two  kinds,  namely,  squares  of  harmonics, 
and  products  of  pairs  of  harmonics.  Let  the  nth  and  the  pth 
harmonics  be  represented  by  the  expressions 

hn  =  Cnsmn(u  -  «„) (71) 

and 

Then  the  right-hand  side  of  eq.  (63)  will  contain  the  following 
terms: 

)  f   hn2dt  =  (CS/T)  C   smzn(u-  an}dt  =  \  Cn2;     (73) 
Jo  Jo 

,j?dt  =  (CP2/D  C  sin2  p(u  -  ap)  dt  =  \  Cp2;      (74) 

2  (l/D  CThnhpdt  = 
Jo 

2  (CnCp/T)  fTs\n  n(u-  an)  sin  p(u  -  ap)  dt  =  0 (75) 

Jo 

The  values  of  the  first  two  integrals  are  found  in  precisely  the 
same  way  as  that  of  eq.  (52)  in  Art.  16,  that  is,  on  the  basis  of 
the  fact  that  their  values  do  not  change  if  cosines  are  substituted 
for  the  sines.  The  third  integral  is  identically  equal  to  zero,  as 
is  shown  in  problem  3  below.  Thus  eq.  (63)  becomes 

y<tf  =  (Ci/V2)»  +  (C,/V2)S  +  etc.,    .     .     .     (76) 

or  the  square  of  the  effective  value  of  a  complex  wave  is  equal  to 
the  sum  of  the  squares  of  the  effective  values  of  its  harmonics. 

Prob.  1.  Plot  a  complex  wave  consisting  of  known  harmonics  and 
determine  its  effective  value  (a)  by  the  method  given  in  the  preceding 
article;  (b)  by  the  Fleming  method;  (c)  from  eq.  (76). 

Prob.  2.  An  irregular  wave  has  a  third  and  a  fifth  harmonic,  the 
amplitudes  of  which  are  equal  respectively  to  12  per  cent  and  4  per  cent 
of  that  of  the  first  harmonic.  Show  that  the  effective  ordinate  is  equal 
to  71.3  per  cent  of  the  amplitude  of  the  fundamental  wave,  and  that  the 
average  value  depends  upon  the  phase  positions  of  the  harmonics. 


CHAP.  V]    POWER  IN  ALTERNATING-CURRENT  CIRCUITS  55 

Prob.  3.  Prove  that  expression  (75)  is  identically  equal  to  zero. 
Proof :  According  to  the  familiar  formula  of  trigonometry,  sin  A  sin  B  = 
%  cos  (A  —  B)  —  \  cos  (A  +  B),  we  have  sin  n(u  —  an)  sin  p(u  —  ap)  = 
%  cos  [(n  —  p}  u  +  a]  —  %  cos  [(n  +  p)  u  +  b],  where  a  and  b  do  not 
contain  the  variable  u.  Integrating  these  cosines  leads  to  terms  of  the 
form  sin  [(n  —  p)  u  +  a]  and  sin  [(n  +  p)  u  +  b].  Since  the  limits  of 
integration  are  0  and  2  w,  and  n  and  p  are  integers,  the  values  of  these 
sines  at  the  upper  limit  are  the  same  as  at  the  lower  limit,  and  conse- 
quently each  of  the  integrals  is  equal  to  zero. 

Vl9.  Power  when  Current  and  Voltage  are  out  of  Phase. 

In  a  majority  of  practical  alternating-current  circuits  there  is  a 
more  or  less  pronounced  phase  displacement  between  the  current 
and  the  voltage.  This  is  due  to  the  presence  of  local  electro- 
motive forces,  the  principal  among  these  being  as  follows:  (a) 
The  counter-electromotive  forces  of  motors  connected  into  the  cir- 
cuit, (b)  The  electromotive  forces  induced  by  alternating  mag- 
netic fluxes  in  the  circuit.  These  fluxes  may  be  created  by  the 
current  itself,  or  they  may  be  due  to  the  influence  of  other  circuits 
(self  and  mutual  induction),  (c)  The  electromotive  forces  due  to 
the  "elastivity"  of  the  dielectric  medium  surrounding  the  circuit 
(electrostatic  capacity  or  permittance). 

The  actual  workings  of  these  causes  are  discussed  more  in 
detail  in  the  following  chapters.  Here  it  is  sufficient  to  note  that 
there  are  factors  which  produce  local  electromotive  forces  in 
alternating-current  circuits,  and  that  they  bring  about  a  phase 
displacement  between  the  voltage  and  the  current.  Let  OB 
(Fig.  13)  be  the  generator  voltage,  and  let  OA  represent  the  sum 
of  the  various  local  electromotive  forces  in  the  circuit.  Sub- 
tracting OA  from  OB,  the  net  voltage  OC  is  obtained,  which  is 
just  sufficient  to  supply  the  ohmic  drop  in  the  circuit.  The 
current  OD  is  in  phase  with  this  voltage,  and  is  numerically  equal 
to  OC  divided  by  the  total  resistance  r  of  the  circuit.  It  will  be 
seen  that  there  is  a  phase  displacement.  <£  between  the  current  and 
the  generator  voltage  OB;  it  is  also  clear  from  the  figure  that  this 
phase  displacement  is  due  to  the  presence  of  the  electromotive 
force  BC. 

We  shall  first  calculate  the  energy  supplied  by  the  generator 
during  one  cycle  in  the  specific  case  when  the  phase  displacement 
between  the  current  and  the  voltage  is  exactly  90  degrees.  If 
the  current  is  represented  by  the  equation  i  =  Im  sin  u,  the 
expression  for  the  voltage  is  e  =  Em  cos  u.  The  instantaneous 


56  THE  ELECTRIC  CIRCUIT  [ART.  19 

power  is  equal  to  i  •  e  =  ImEm  sin  u  cos  u  =  \  ImEm  sin  2  u.  Thus, 
the  power  varies  as  a  sine  function  of  double  the  generator  fre- 
quency; the  energy  flows  now  away  from,  and  now  toward, 
the  generator.  The  average  power  for  one  cycle  is  therefore  zero, 
for  the  power  has  as  many  negative  values  as  it  has  positive  ones. 
Mathematically,  this  result  is  represented  by  the  time  integral 
of  the  instantaneous  power  over  a  complete  cycle.  Omitting  the 
constant  quantities  Em  and  Im,  we  have 

/     smucosudu  =  \   —  cos2w      =0. 

Let  now  the  phase  displacement  between  the  current  and  the 
voltage  be  less  than  90  degrees,  and  be  equal,  say,  to  <£.  The 
average  power  delivered  by  the  alternator  is  in  this  case  smaller 
than  the  product  El,  and  its  value  must  be  investigated.  The 
vector  of  the  voltage  E  can  be  resolved  into  a  component  E  cos  <£ 
in  phase  with  the  current,  and  another  component,  E  sin  0,  in 
quadrature  with  the  current.  According  to  the  proof  given 
above,  the  average  power  produced  by  the  quadrature  component 
of  the  voltage  is  zero,  so  that  the  total  average  power  is 

Pave  =  El  -  cos  0 (77) 

A  more  rigid  proof  of  this  expression  is  given  in  problem  3  below. 

The  product  El  is  called  the  apparent  power,  and  cos<£  is 
referred  to  as  the  power-factor.  Thus,  the  power-factor  can  be 
defined  either  as  the  cosine  of  the  angle  of  phase  displacement 
between  the  current  and  the  voltage,  or  as  the  ratio  of  the  true 
power  Pave  to  the  apparent  power  IE.  The  second  definition 
is  more  general,  because  it  applies  also  to  non-sinusoidal  currents 
and  voltages. 

Referring  to  Fig.  13,  the  factor  7  cos  <£  which  enters  into 
eq.  (77)  represents  the  projection  of  7  upon  the  direction  of  the 
voltage  OB,  or  E.  Hence,  eq.  (77)  can  be  interpreted  by  saying 
that  the  true  power  is  equal  to  the  product  of  the  voltage  by 
the  component  of  the  current  in  phase  with  it.  This  component 
of  the  current,  I  cos  <£,  is  therefore  called  the  energy  component, 
while  the  component  7  sin  <j>,  at  right  angles  or  in  quadrature  with 
the  voltage,  is  called  the  reactive  component.1 

Instead  of  resolving  the  vector  of  the  current  into  two  com- 
ponents, it  is  sometimes  preferable  to  resolve  the  voltage  E  into 
1  The  older  name  for  this  reactive  component  is  wattless  current. 


CHAP.  V]    POWER  IN  ALTERNATING-CURRENT  CIRCUITS          57 

the  components  E  cos  <£  and  E  sin  <£,  in  phase  and  in  quadrature 
with  the  current.  In  this  case,  eq.  (77)  is  expressed  in  words  by 
saying  that  the  average  power  is  equal  to  the  current  times  the 
component  of  the  voltage  in  phase  with  it.  These  components 
of  the  voltage  are  also  called  the  energy  component  and  the  reactive 
component  respectively.  The  two  components  of  power,  the  true 
power  El  cos  </>,  and  the  reactive  power  El  sin  0,  stand  in  the 
same  relation  to  the  apparent  power  El  as  the  two  sides  of  a 
right  triangle  bear  to  the  hypotenuse;  that  is, 

(El)*  =  (El  cos  <£)2  +  (El  sin  </>)2 (78) 

Let  now  the  current  and  voltage  curves  be  different  from  pure 
sine-waves,  and  also  different  from  each  other  in  form.  The 
fundamental  equation 

Pa*e=  (1/T)  fT ei-dt  (79) 

Jo 

holds  true  in  all  cases,  so  that  if  the  curves  are  given  graphically, 
the  energy  per  cycle  is  found  by  multiplying  the  corresponding 
instantaneous  values  of  e  and  i,  and  using  the  planimeter  on  the 
resultant  curve.  The  average  ordinate  of  this  curve  gives  the 
average  power.  Of  course,  the  parts  of  the  resultant  curve  below 
the  axis  of  abscissae  must  be  evaluated  separately  from  those  above 
it,  and  the  difference  of  the  two  taken  to  represent  the  total 
energy. 

If  the  two  waves  are  given  in  the  form  of  Fourier  series,  an 
expression  for  the  average  power  may  be  obtained  in  terms  of 
the  effective  values  of  the  harmonics.  Substituting  the  expan- 
sions for  e  and  i  into  eq.  (79),  two  kinds  of  terms  are  obtained, — 
those  containing  products  of  two  harmonics  of  the  same  frequency, 
and  those  containing  products  of  two  harmonics  of  different  fre- 
quencies. The  terms  of  the  first  kind,  after  integration,  give  re- 
sults of  the  same  form  as  for  the  fundamental  wave;  that  is,  for 
the  nth  harmonic  \  Enln  cos  <£„,  where  En  and  /„  are  the  amplitudes 
of  the  nth  harmonics,  and  <f>n  is  the  phase  displacement  between 
them.  The  terms  of  the  second  kind  give  zero  after  integration, 
the  proof  of  this  being  analogous  to  that  in  problem  3  of  the  pre- 
ceding article.  Thus 

Pave  =  vEJi  cos  0i  +  \EJZ  cos  <£3  +  etc.  .  .  (80) 
In  other  words,  each  harmonic  contributes  its  own  share  of  power, 
as  if  it  were  acting  alone. 


58  THE  ELECTRIC  CIRCUIT  [ART.  19 

Let  E  and  7  be  the  effective  values  of  some  non-sinusoidal 
periodic  voltage  and  current,  measured,  for  instance,  by  means 
of  hot-wire  or  dynamometer-type  instruments.  Let  Pave  be  the 
average  power  according  to  eq.  (80),  or  measured  by  a  dyna- 
mometer-type wattmeter.  Then  the  ratio  Pave/EI  is  called  the 
power-factor  of  the  system,  the  same  as  with  sinusoidal  curves. 
This  ratio  is  also  often  denoted  by  cos  <£,  meaning  by  </>  the  phase 
angle  between  the  equivalent  sine-waves  of  voltage  and  current, 
as  defined  in  the  preceding  article.  With  the  use  of  this  angle  and 
of  the  equivalent  sine-waves,  vector  diagrams  may  be  constructed 
and  the  corresponding  calculations  performed  with  currents  and 
voltages  deviating  considerably  from  pure  sine-waves,  though  of 
course  such  calculations  check  only  approximately  with  the  actual 
measurements. 

Prob.  1.  Assuming  the  line  current  in  problem  4,  Art.  14,  to  be  452 
effective  amperes,  calculate  the  average  power  delivered  by  the  alternator, 
and  the  power  received  at  the  opposite  end  of  the  line. 

Ans.     2444  kw.;  2350  kw. 

Prob.  2.  Referring  to  problem  1,  Art.  17,  a  wattmeter  was  connected 
into  the  heater  circuit,  and  the  true  power  was  found  to  be  598  watts. 
Assuming  all  the  three  instruments  to  be  in  calibration,  calculate  the 
power-factor  and  the  angle  of  displacement  between  the  current  and  the 
voltage  in  the  heater;  also  the  energy  component  and  the  reactive  com- 
ponent of  the  current. 

Ans.     95.4  per  cent;  17°  30';  5.39  amp.;  171  amp. 

Prob.  3.  Deduce  expression  (77)  for  power  by  direct  integration. 
Solution:  Let  the  current  be  expressed  by  /,„  sin  u;  also  let  the  voltage 
he  leading  by  an  angle  <t>,  and  therefore  expressed  as  Em  sin  (u  +  <#>). 
Substituting  these  values  into  eq.  (79),  we  get 

CT 
l\ve  =  (E,,Jm/T)  I     sin  u  sin  (u  +  $)  dt, 

Jo 

=  (EmIm/2  *•)  /      sin  u  [sin  u  cos  <j>  +  cos  u  sin  <f>]  du, 

=  (EmIm/2  *•)  [cos  tt>  I      sin2  u  du  +  sin  <£  /      sin  u  cos  udu]. 
Jo  Jo 

From  a  table  of  integrals  we  find  that  the  value  of  the  first  integral  is  *-, 
and  that  of  the  second  is  zero.  Substituting  these  values,  and  introduc- 
ing the  effective  values  of  voltage  and  current,  formula  (77)  is  obtained. 
Prob.  4.  Plot  a  sine-wave  representing  an  alternating  voltage  of 
500  effective  volts,  and  a  current  of  the  same  frequency,  of  20  effective 
amperes,  lagging  behind  the  voltage  by  30  degrees.  Plot  on  the  same 
curve  sheet  the  sine-wave  of  the  instantaneous  power,  and  check  the 


CHAP.  V]    POWER  IN  ALTERNATING-CURRENT  CIRCUITS  59 

average  ordinate  of  this  curve  with  the  value  obtained  by  formula  (77). 
Explain  why  the  power  is  negative  during  a  part  of  the  cycle,  remember- 
ing that  there  are  local  electromotive  forces  in  the  circuit. 

Prob.  5.  Prove  that  the  curve  of  power  consists  of  a  sine- wave  of 
double  frequency,  plus  a  constant  term,  the  latter  representing  the  aver- 
age power.  Compare  with  problem  3,  Art.  16.  Suggestion:  ie  =  ImEm 
sin  u  sin  (u  +  <f>}.  Use  the  trigonometric  transformation,  2  sin  A  sin  B  = 
cos  (A  -  B]  -  cos  (A  +  B). 

Prob.  6.  A  non-sinusoidal  voltage  is  represented  by  the  equation 
e  =  270  sin  u  +  62  sin  3(u  +  15°)  +  16  sin  5(u  -  25°);  the  correspond- 
ing line  current  is  i  =  18  sin  (u  -  30°)  -  7  sin  3  (u  +  50°)  +  2.5  sin 
5(w  +  10°).  Calculate  the  true  average  power  and  the  power-factor  of 
the  system. 

Ans.  Pave  =  \  (4860  cos  30°  -  454  cos  75°  -  40  cos  5°)  =  2027  watte; 
the  power-factor  is  75  per  cent. 


CHAPTER  VI 
INDUCTANCE,  REACTANCE  AND  IMPEDANCE 

20.  Inductance  as  Electromagnetic  Inertia.  Experiment 
shows  that  an  electric  current  in  a  variable  state  behaves  as  if 
it  possessed  inertia;  there  is  an  opposition  to  any  change  in  its 
magnitude  and  direction.  This  opposition  is  manifested  in  the 
form  of  an  "induced"  electromotive  force  in  such  a  direction  as 
to  tend  to  counteract  the  change  in  current.  Thus,  if  an  external 
e.m.f.  tends  to  increase  the  current,  the  induced  e.m.f.  is  in  a 
direction  opposite  to  that  of  the  current;  but  when,  on  the  other 
hand,  the  current  for  some  reason  decreases,  the  induced  e.m.f. 
is  in  the  same  direction  as  the  current,  and  therefore  tends  to 
strengthen  it.  These  reactions  of  the  current  are  similar  to  those 
exerted  by  a  moving  body;  for  instance,  the  water  in  a  pipe,  when 
its  motion  is  accelerated  or  retarded.  In  practical  applications,  it 
is  convenient  to  consider,  not  the  reactions  themselves,  but  the 
external  forces  necessary  to  overcome  them.  Thus,  in  the  case 
of  a  moving  body  of  mass  m,  the  external  force  necessary  to  com- 
municate to  it  an  acceleration  dv/dt  is  F  =  m  dv/dt.  Here  F 
is  positive  when  the  acceleration  is  positive,  and  vice  versa. 
Similarly,  to  increase  a  current  at  a  rate  of  di/dt,  an  external  e.m.f. 
is  necessary  of  the  magnitude 

e  =  Ldi/dt, (81) 

where  L  is  a  constant  which  characterizes  the  circuit  and  is  analo- 
gous to  the  mass  m  in  the  mechanical  motion.  The  coefficient  L 
is  called  the  inductance  of  the  circuit,  and  depends  upon  its  shape 
and  proportions,  the  presence  or  absence  of  iron,  the  number  of 
turns  which  the  conductor  makes,  and  some  other  factors,  which 
it  is  not  necessary  to  discuss  here.  In  most  books  the  right-hand 
side  of  eq.  (81)  is  written  with  the  sign  minus,  because  e  is  under- 
stood to  mean  the  induced  e.m.f.  or  the  reaction  of  the  circuit; 
while  in  our  case  e  designates  the  external  voltage,  equal  and 

60 


CHAP.  VI]       INDUCTANCE,  REACTANCE  AND  IMPEDANCE        61 

opposite  to  this  reaction.  The  form  of  the  equation  used  here  is 
preferable,  because  in  practice  one  deals  with  components  of  the 
applied  voltage  rather  than  with  the  induced  counter-e.m.f . ; 
moreover,  the  minus  sign  is  apt  to  confuse  a  beginner. 

The  inertia  effect  of  the  electric  current  is  brought  about 
through  the  mechanism  of  the  magnetic  field  produced  thereby. 
When  the  current  varies,  the  flux  embraced  by  the  electric  circuit 
also  changes,  and  according  to  Faraday's  law  of  induction  this 
flux  induces  in  the  circuit  an  e.m.f.  Thus,  postulating  the  exist- 
ence of  electromagnetic  inertia,  and  stating  the  law  of  induced 
e.m.f.,  are  perhaps  but  two  different  ways  of  expressing  the  same 
physical  phenomenon,  the  true  nature  of  which  is  at  present  un- 
known. Any  arrangement  of  the  circuit  which  increases  the  flux 
linked  with  it,  also  increases  its  inductance  L  or  the  inertia  effect. 
The  inductance  of  a  given  electric  circuit  can  be  calculated  with 
more  or  less  accuracy,1  or  it  can  be  measured  experimentally, 
using  eq.  (81).  For  our  present  purposes  we  shall  assume  L  to 
be  a  constant  quantity,  which  characterizes  the  inertia  of  a  given 
electric  circuit,  according  to  eq.  (81),  without  any  reference  to 
the  nature  of  the  magnetic  flux  which  produces  it.  Mechanical 
inertia  is  used  in  physics  and  in  engineering  as  a  fundamental 
entity,  without  explaining  it  in  any  other  terms,  while  the  mystery 
as  to  its  cause  is  just  as  deep  as  that  surrounding  the  electro- 
magnetic inertia.  Some  modern  physicists  even  believe  that  all 
inertia  is  of  an  electromagnetic  nature. 

The  fact  that  a  body  resists  acceleration,  together  with  the 
law  of  conservation  of  energy,  leads  to  the  conclusion  that  a  mov- 
ing body  possesses  a  certain  amount  of  stored  energy.  The 
external  work  done  upon  a  body  while  it  moves  through  a  dis- 
tance ds  is  F  •  ds  =  m  (dv/dt)  ds,  or,  since  ds  =  v  •  dt,  we  have 
F  •  ds  =  mv  •  dv.  The  total  work  done  upon  the  body  while  accel- 
erating it  from  rest  to  a  velocity  v  is  therefore 

W  =  CS  Fds  =  r  mv  •  dv  =  \  mv2. 

According  to  the  law  of  conservation  of  energy,  this  work  is 
stored  in  the  moving  body  as  its  kinetic  energy. 

"  The  electrical  work  done  in  increasing  a  current  against  the 
induced  electromotive  force,  during  the  time  dt  is  d W  =  ei  •  dt,  or 
1  See  the  author's  Magnetic  Circuit,  Chapters  10  to  12, 


62  THE   ELECTRIC  CIRCUIT  [ART.  20 

substituting  for  e  its  value  from  eq.  (81),  dW=  Lidi.  The  total 
energy  supplied  to  the  circuit  from  the  external  source  of  power, 
while  the  current  increases  from  zero  to  a  certain  value  i,  is 

W  =  r  Lidi  =  $Li* (82) 

Jo 

This  does  not  include  the  energy  required  for  supplying  the  i-r 
loss.  According  to  the  law  of  conservation  of  energy,  expression 
(82)  represents  the  energy  stored  in  the  circuit  as  long  as  the  value 
of  the  current  remains  the  same.  When  the  circuit  is  broken, 
this  energy  is  converted  into  heat.  Analogously,  when  a  non- 
elastic  moving  body  is  stopped,  its  accumulated  energy  is  con- 
verted into  the  heat  of  impact.  Inductance  can  be  defined  from 
either  eq.  (81)  or  (82);  and  for  most  purposes  the  two  definitions 
are  identical.  Similarly,  in  mechanics,  mass  may  be  defined 
either  as  the  ratio  of  F  to  dv/dt,  or  as  a  ratio  of  the  kinetic  energy 
to  \vz. 

The  unit  of  inductance  in  the  ampere-ohm  system  is  called 
the  henry.  According  to  eq.  (81),  a  circuit  has  an  inductance  of 
one  henry  when  one  volt  is  necessary  in  order  to  increase  the 
current  at  a  rate  of  one  ampere  per  second.  This  one  volt  does 
not  include,  of  course,  the  e.m.f.  necessary  for  overcoming  the 
resistance  of  the  circuit.  The  henry  being  rather  a  large  unit, 
inductance  is  frequently  measured  in  millihenrys.  Substituting 
into  eq.  (81)  the  physical  dimensions  of  the  voltage  in  the  ampere- 
ohm  system,  we  get  [IR]  =  [LI/T]  or  [L]  =  [RT].  In  other 
words,  the  henry  stands  for  the  "ohm-second."  For  this  reason, 
one  instrument  for  measuring  inductance  has  been  called  by  its 
inventors  "the  secohmmeter. " 

All  actual  circuits  which  possess  inductance,  at  the  same  time, 
have  some  resistance,  however  small  it  may  be.  Therefore,  the 
total  instantaneous  voltage  applied  during  a  variable  state  is 

e  =  ir  +  L  di/dt (83) 

Ohmic  resistance  may  be  compared  to  mechanical  friction, 
so  that  eq.  (83)  can  be  interpreted  by  reference  to  the  mechan- 
ical analogy  used  above,  in  the  following  way;  namely,  the 
force  necessary  to  accelerate  a  body  must  be  augmented  in 
practice  by  the  amount  required  for  overcoming  the  inevitable 
friction. 


CHAP.  VI]        INDUCTANCE,  REACTANCE  AND  IMPEDANCE       63 

Prob.  1.  A  circuit  which  possesses  an  inductance  of  12  millihenrys 
and  carries  a  direct  current  of  150  amp.  is  broken  within  one-fifth  of  a 
second.  What  is  the  average  voltage  induced  in  the  circuit  during  this 
interval  of  time?  Ans.  9  volts. 

Prob.  2.  Calculate  the  electromagnetic  energy  stored  in  the  circuit  of 
the  preceding  problem  while  the  current  is  steady.  Ans.  135  joules. 

Prob.  3.  The  current  in  a  coil  is  made  to  vary  at  a  uniform  rate  of 
250  amp.  per  second.  At  the  instant  when  the  current  is  equal  to  150 
amp.,  a  voltmeter  connected  across  the  terminals  of  the  coil  reads  295 
volts;  when  the  instantaneous  current  is  100  amp.  the  voltmeter  reading 
is  230  volts.  From  these  data  calculate  the  resistance  and  the  induc- 
tance of  the  coil.  Ans.  1.3  ohms;  0.4  henry. 

21.  Reactance.  It  is  natural  to  expect  the  inductance  to 
exert  a  considerable  influence  upon  the  voltage  and  current  rela- 
tions in  an  alternating-current  circuit,  because  the  Current  is 
varying  in  magnitude  all  the  time.  The  influence  of  inductance 
in  this  case  is  analogous  to  that  of  the  inertia  of  the  moving  parts 
in  a  reciprocating  engine;  i.e.,  energy  is  stored  during  the  periods 
of  increase  in  velocity  (or  in  current),  and  is  returned  to  the  source 
of  power  during  the  intervals  of  time  when  the  velocity  (or  the 
current)  decreases.  There  is  no  net  gain  or  loss  of  energy  for  a 
complete  cycle,  although  the  instantaneous  values  of  current  and 
voltage  may  be  considerably  affected. 

Consider  first  a  part  of  a  circuit  which  has  inductance  only, 
the  resistance  being  negligible.  Let  the  current  vary  according 
to  the  familiar  law  i  =  Im  sin  (2  irft  —  a).  Substituting  this  value 
into  eq.  (81),  we  get 

e  =  27r/L7mcos(27r/Y-«),       ....     (84) 

which  means  that  the  voltage  necessary  to  force  a  sinusoidal 
current  through  an  inductance  also  varies  according  to  the  sine 
law,  and  is  in  leading  quadrature  with  the  current.  The  ampli- 
tude of  the  voltage  Em  =  2  -irfLIm,  or  the  relation  between  the 
effective  values  of  voltage  and  current,  is 

E  =  2irfLI (85) 

It  will  be  seen  from  this  relation  that,  in  alternating-current  cal- 
culations, the  quantities  /  and  L  always  appear  as  a  product. 
It  is  therefore  convenient  to  introduce,  for  the  sake  of  abbrevia- 
tion, a  new  composite  quantity  x,  defined  by  the  relation 

.     .  (86) 


64 


THE  ELECTRIC  CIRCUIT 


[ART.  21 


The  quantity  x  is  called  the  reactance  of  the  circuit,  and  always 
refers  to  a  stated  frequency  /.  Equation  (85)  becomes  then 

E  =  xl, (87) 

from  which  it  follows  that  reactance  is  measured  in  ohms,  like 
resistance.  This  does  not  mean,  however,  that  the  two  quanti- 
ties are  similar  in  their  physical  nature. 

r  Let  now  some   resistance  be  con- 

.       nected  in  series  with  the  inductance 
E  ||l     (Fig.   18),  or  let  the  coil  which  pos- 

sesses inductance  have  also  an  appre- 
ciable resistance.  Substituting  the 
expression  for  i,  given  above,  into 
eq.  (83),  we  get 

e  =  rlm  sin  (2  irft  -  a)  +  xlm  cos  (2  */«-«)..     .     (88) 

Since  the  sum  of  two  sine-waves  is  also  a  sine-wave  (see  Art.  14), 
the  total  voltage  e  varies  according  to  the  sine-law.  These  com- 
ponent sine-waves  of  voltage,  their  sum,  and  the  current  wave 
are  shown  in  Fig.  19.  The  student  is  advised  to  study  this  figure 


FIG.  18.     Resistance  and  re- 
actance in  series. 


FIG.  19.    The  instantaneous  current  and  voltage  relations  in  the  circuit 
shown  in  Fig.  18. 

very  carefully,  because  it  represents  one  of  the  most  important 
fundamental  relations  in  the  whole  theory  of  alternating  currents. 
The  same  relations  are  represented  vectorially  in  Fig.  20,  and  the 
two  figures  may  be  conveniently  examined  together.  One  de- 
scribes the  phenomenon  from  instant  to  instant;  the  other  gives 


CHAP.  VI]       INDUCTANCE,  REACTANCE  AND  IMPEDANCE         65 

the  salient  features  in  a  symbolic  form.  The  vector  E  consists 
of  one  component  Ir  in  phase  with  the  current,  and  another  Ix 
in  leading  quadrature  with  the  current.  The  first  component 


ir         i 

FIG.  20.    The  current  and  voltage  relations  in  the  circuit  shown  in  Fig.  18, 
represented  vectorially. 

serves  to  overcome  the  ohmic  resistance;  the  second,  the  reac- 
tance of  the  circuit.     From  the  triangle  of  voltages  we  have 


(89) 

For  the  phase  displacement  between  the  current  and  the  voltage 
we  have 

tan<£  =  Ix/Ir  =  x/r,  ......     (90) 

or  the  power-factor 

cos</>  =  r/Vr2  +  z2  .......     (91) 

The  hydraulic  analogue  shown  in  Fig.  21  may  make  these 
relations  clearer.  ACDGA  represents  a  closed  pipe  circuit  in 
which  water  is  made  to  oscillate 
to  and  fro  by  means  of  the  piston 
B.  The  water  is  assumed  to  be 
devoid  of  inertia,  and  the  inertia 
of  the  whole  circuit  is  concen- 
trated in  a  heavy  mass  F,  which 
moves  freely  with  the  water. 
The  force  upon  the  piston  rod 
H  is  analogous  to  the  alternat- 
ing voltage  E  in  Fig.  18;  the 
velocity  of  the  water  is  analo- 


FIG. 21.    A  hydraulic  analogue 
to  Fig.  18. 


gous  to  the  alternating  current,  the  friction  in  the  pipes  represents 
the  ohmic  resistance  r,  and  the  inertia  of  the  heavy  mass  F  stands 
for  the  inductance  L.  To  make  the  analogy  closer,  we  assume 
that  the  piston  is  forced  to  perform  a  simple  harmonic  motion; 


66  THE  ELECTRIC  CIRCUIT  [ART.  21 

so  that  the  velocity  of  the  water  varies  with  the  time  according 
to  the  sine  law,  and  may  be  represented  by  the  curve  for  i  in 
Fig.  19. 

The  force  upon  the  piston  B  consists  of  two  parts,  that  re- 
quired for  overcoming  the  friction  in  the  pipes,  and  that  necessary 
for  accelerating  and  retarding  the  mass  F.  These  two  compo- 
nents of  the  force  can  be  represented  by  the  curves  for  ir  and  ix 
in  Fig.  19.  The  frictional  reaction  is  at  a  maximum  when  the 
piston  is  in  the  middle  of  its  stroke,  because  there  the  velocity 
of  the  water  is  the  greatest.  On  the  other  hand,  the  acceleration 
is  zero  in  this  position,  so  that  the  mass  F  exerts  no  reaction. 
At  the  ends  of  the  stroke  the  acceleration  or  retardation  is  at  a 
maximum,  so  that  the  force  necessary  for  constraining  the  mass 
F  to  the  prescribed  motion  is  at  a  maximum;  however,  the  fric- 
tional resistance  is  equal  to  zero.  Adding  the  two  sinusoidal  com- 
ponents, we  find  the  resultant  force  upon  B,  corresponding  to 
the  curve  e  in  Fig.  19.  It  will  be  seen  that  e  reaches  a  maximum 
before  the  center  of  the  stroke;  this  gives  a  phase  angle  between 
the  force  and  the  velocity  that  is  analogous  to  the  phase  angle 
between  the  voltage  and  the  current.  The  student  can  easily 
deduce  that  the  force  leads  the  velocity  in  phase,  and  that  the 
displacement  is  greater  the  larger  the  mass  F,  as  compared  to 
the  frictional  resistance;  in  other  words,  the  greater  the  reactance 
as  compared  to  the  resistance.  It  may  be  shown  also  that  the 
inertia  reaction  of  the  same  mass  F  is  greater  for  a  higher  fre- 
quency of  oscillation,  because  the  acceleration  and  retardation  are 
proportionately  larger. 

Prob.  1.  The  inductance  of  a  coil  is  0.2  henry;  its  ohmic  resistance 
is  negligible.  Draw  a  curve  giving  the  voltage  necessary  to  maintain  a 
current  of  12  amp.  through  the  coil,  at  frequencies  ranging  from  zero  to 
100  cycles  per  second. 

Ans.    A  straight  line  through  the  origin;  at  /  =  100,  E  =  1508  volts. 

Prob.  2.  A  reactive  coil  without  iron  draws  a  current  of  75  amp.  when 
connected  across  a  110- volt  25-cycle  circuit.  What  current  would  it  draw 
at  60  cycles  and  at  the  same  voltage,  provided  that  the  effect  of  its 
resistance  can  be  neglected?  Plot  a  curve  of  current  at  intermediate 
frequencies. 

Ans.    31.25  amp.;  equilateral  hyperbola  asymptotic  to  both  axes. 

Prob.  3.  The  reactive  magnetizing  current  of  a  2200-volt,  600-kilo- 
volt-ampere,  50-cycle  transformer  must  be  not  over  2.5  per  cent  of  the 
full-load  current.  What  is  the  lower  limit  of  its  no-load  reactance  and 
inductance?  Ans.  322.5  ohms;  1.027  henrys. 


CHAP.  VI]        INDUCTANCE,  REACTANCE  AND  IMPEDANCE       67 

Prob.  4.  The  coil  considered  in  problem  1  is  connected  in  series  with 
a  100-ohm  resistance;  it  is  required  to  maintain  a  current  of  12  amp. 
through  the  two,  at  various  frequencies.  Supplement  the  curve  obtained 
in  that  problem  with  curves  of  voltage  drop  across  the  resistance,  and  the 
total  voltage  across  the  combination.  Plot  also  the  corresponding  values- 
of  power-factor.  Determine  the  ordinates  of  the  curves  graphically, 
and  check  a  few  points  analytically. 

Ans.  Er  =  1200  volts,  independent  of  the  frequency.  At  /  =  0, 
E Mai  =Er,  and  cos  0=1.  At  /=  100,  Etotai  =  1927  volts,  cos  0  =  62.25 
per  cent. 

Prob.  5.  Three  simultaneous  instrument  readings  in  a  power  house 
are:  7520  kw.;  66  kv.;  147  amp.  The  power-factor  meter  shows  that 
the  current  is  lagging  behind  the  voltage.  What  are  the  readings  at  the 
same  instant  at  the  receiving  end  of  the  line,  if  its  resistance  is  45  ohms 
and  its  reactance  83  ohms.  Hint:  Draw  the  vectors  of  the  generator 
voltage  and  current  in  their  true  relative  position.  Subtract  the  ohmic 
drop  in  phase  with  the  current,  and  the  reactive  drop  in  quadrature  with 
it.  The  result  will  give  the  receiver  voltage  in  its  true  magnitude  and 
phase  position.  Ans.  6547  kw.;  53.4  kv. 

Prob.  6.  In  order  to  determine  the  power  input  into  a  single-phase 
110- volt  motor,  without  the  use  of  a  wattmeter,  the  motor  is  connected 
in  series  with  a  non-inductive  resistance  across  a  220-volt  circuit.  The 
resistance  is  adjusted  so  that  the  voltage  across  the  motor  terminals  is 
110,  when  the  motor  is  carrying  the  required  load.  Under  these  condi- 
tions the  voltage  across  the  resistance  is  found  to  be  127,  and  the  current 
through  the  motor  23  amp.  From  these  data  determine  graphically  the 
power-factor  of  the  motor,  and  calculate  its  power  input. 

Ans.    72.3  per  cent;  1826  watts. 

Prob.  7.  Referring  to  the  preceding  problem,  calculate  cos  <f>  trigo- 
nometrically,  from  the  triangle  of  voltages,  instead  of  determining  it 
graphically. 

22.  Impedance.  When  a  reactance  is  connected  in  series 
with  a  resistance,  eqs.  (89)  and  (91)  indicate  that  the  current  and 
voltage  relations  are  determined,  not  by  the  value  of  the  reac- 
tance alone,  but  by  a  composite  expression 

z  =  Vr*  +  x\ (92) 

The  quantity  z  has  the  dimension  of  a  resistance,  and  is  called 
the  impedance  of  the  circuit.  It  can  hardly  be  called  a  physical 
quantity,  but  rather  an  abbreviation  for  a  certain  combination 
of  the  physical  properties  of  a  circuit;  in  other  words,  an  abbre- 
viation for  the  radical  in  eq.  (92).  Introducing  the  value  of  z 
into  eqs.  (89)  and  (91),  we  obtain 

E  =  zl (93) 

and  cos  0  =  r/z.  . (94) 


THE  ELECTRIC  CIRCUIT 


[ART.  22 


Impedance  may  be  defined  from  eq.  (93)  as  the  ratio  of  the  voltage 
to  the  current  in  a  circuit  containing  resistance  and  reactance. 
In  a  non-inductive  circuit  the  impedance  is  simply  equal  to  the 
total  resistance,  while  in  a  purely  inductive  one  the  impedance 
is  equal  to  the  reactance.  It  must  be  clearly  understood  that 
eq.  (93)  gives  only  the  relation  between  the  magnitudes  of  the 
vectors.  The  phase  relation  is  given  by  Fig.  20,  or  by  eq.  (94). 
The  three  quantities  r,  x,  and  z  form  a  triangle  of  which  z  is 
the  hypothenuse  (Fig.  20).  This  triangle  is  similar  to  the  tri- 
angle of  voltages,  but  the  quantities  r,  x,  and  z  are  not  vectors  in 
the  same  sense  as  currents  and  voltages  are.  From  the  impe- 
dance triangle  we  have  the  following  useful  relations: 

r  =  z  cos  0 (95) 

and 

x  =  z  sin  </> (96) 

When  two  impedances  are  connected  in  series  (Fig.  22),  the 
voltage  and  current  relations  are  as  represented  in  Fig.  23.     The 

total  terminal  voltage  E  is  less 
than  the  arithmetical  sum  of 
the  voltages  EI  and  E2  across 
the  two  impedances,  and  is 
equal  to  their  geometric  sum. 
FIG.  22.  Two  impedances  in  series.  The  resultant  phase  angle  0  has 

a  value  intermediate  between 

the  phase  angles  <f>i  and  <£2  of  the  two  component  impedances. 
It  will  be  seen  from  the  triangle  ABC  that  the  resultant  voltage 


Fia.  23.     The  current  and  voltage  relations  in  the  circuit  shown  in  Fig.  22. 

is  the  same  as  that  required  by  an  impedance  which  consists 
of  a  resistance  TI  +  r2  and  a  reactance  Xi  +  z2.     In  other  words, 


CHAP.  VI]        INDUCTANCE,  REACTANCE  AND  IMPEDANCE        69 

the  resultant  impedance  is 


z  =      (n  +  r2)2  +  (Xl  +  z2)2,      ....     (97) 
and  the  resultant  phase  angle  is  determined  from  the  equation 

tan  0  =  (si  +  z2)/(ri  +  r2)  .....  (98) 
These  equations  show  that  two  impedances  are  added  in  series 
by  adding  the  resistances  and  the  reactances  separately.  An 
impedance  of  5  ohms  in  series  with  one  of  7  ohms  is  not  equal  to 
an  impedance  of  12  ohms,  but  as  a  rule  is  less.  The  relations 
shown  in  Figs.  22  and  23,  and  eqs.  (97)  and  (98),  are  easily  ex- 
tended to  any  number  of  impedances  in  series.  Dividing  all  the 
voltage  vectors  in  Fig.  23  by  the  value  of  the  current  7,  the  dia- 
gram of  voltages  is  converted  into  one  of  impedances,  as  in  Fig.  20, 
the  relations  being  represented  by  eqs.  (97)  and  (98).  It  must  be 
borne  in  mind,  however,  that  from  a  physical  point  of  view  the 
latter  relations  are  not  vectorial  in  the  same  sense  as  are  those 
of  the  voltages. 

Prob.  1.  The  impedance  of  a  coil  is  7.5  ohms  at  60  cycles;  the  re- 
sistance measured  with  direct  current  is  6  ohms.  What  is  the  inductance? 

Ans.     11.9  millihenrys. 

Prob.  2.  Two  impedance  coils  are  connected  in  series  across  a  292- 
volt  line.  The  voltages  across  the  coils  are  152  and  175  respectively; 
the  current  is  7.3  amp.  Knowing  that  the  resistance  of  the  first  coil  is 
10  ohms,  determine  graphically  the  resistance  of  the  second;  also  the 
impedances  of  both  coils. 

Ans.    rz  =  23.8;     z\  =  20.82;    z2  =  23.97,  all  in  ohms. 

Prob.  3.  When  a  certain  non-inductive  resistance  is  connected  across 
a  source  of  alternating  voltage,  a  current  /  flows  through  it.  When  an 
inductance,  containing  negligible  resistance,  is  connected  across  the  same 
source  of  voltage,  the  current  is  /'.  What  are  the  current  and  the  phase 
displacement  when  the  resistance  and  the  inductance  are  connected  in 
series  across  the  same  source?  Solution:  Let  the  unknown  voltage  be 
E.  The  unknown  resistance  is  r  =  E/I;  the  unknown  reactance  x  — 
E/I'.  When  the  two  are  connected  in  series,  the  impedance  z  =  [(E/I)2 
+  (E/I'Y\*.  Consequently,  the  current  is  E/z  =  7/'/(/2  +  //2)*  ;  tan  0  = 
x/r  =  I  /I'. 

23.  Influence  of  Inductance  with  Non-sinusoidal  Voltage. 

(a)  Let  an  alternating  voltage  e  of  an  irregular  form,  such  as  is 
shown  in  Fig.  14,  be  applied  at  the  terminals  of  a  pure  resistance  r 
(non-inductive).  The  current  through  the  resistance  is  at  any 
instant  equal  to  e/r,  and  consequently  has  the  same  wave  form 
as  the  voltage. 


70  THE  ELECTRIC  CIRCUIT  [Aux.  23 

(I))  Let  now  the  same  voltage  be  applied  at  the  terminals  of 
a  pure  inductance  L  (without  resistance).  It  may  be  said  a  priori 
that  the  current  wave  will  be  different  from  that  of  the  voltage, 
and  will  approach  more  nearly  a  sine-wave.  This  follows  from  the 
very  concept  of  inductance  as  the  inertia  of  the  circuit;  the  high- 
frequency  harmonics  in  the  voltage  are  unable  to  produce  currents 
of  the  same  magnitude  as  at  lower  frequencies,  because  the  react- 
ance offered  to  each  harmonic  is  proportional  to  its  frequency. 
This  property  of  an  inductance  of  choking  higher  harmonics  is 
useful  in  some  applications. 

Let  the  voltage  across  an  inductance  be  given  in  the  form  of  a 
Fourier  scries, 

e  =  Et  sin  (2  irft  -  «i)  +  #3  sin  3  (2  irft  -  «3)  +  etc. 
Substituting  its  value  in  the  fundamental  eq.  (81),  we  get 

El  sin  (2  Trft  -  «0  +  E3  sin  3  (2  Trft  -  «3)  +  etc.  =  L  di/dt. 
Multiplying  both  sides  of  this  equation  by  dt  and  integrating 
gives 

-  (#!/2  TT/)  COS  (2  Trft  -  «i)  -  (#3/6  TT/)  COS  3  (2  Trft  -  a3) 
—  etc.  =  Li  +  const. 

The  constant  of  integration  is  equal  to  zero,  because  the  current 
cannot  have  a  unidirectional  component  without  a  commutating 
device  or  electric  valve  of  some  sort.     Therefore 
i  =  -  (Ei/2  7T/L)  cos  (2  Trft  -  «0  -  (#3/6  7T/L)  cos  3  (2  Trft  -  «3) 

-  etc., (99) 

which  means  that  each  harmonic  in  the  e.m.f.  produces  its  own 
current,  as  if  this  harmonic  were  acting  alone.  The  total  current 
is  the  sum  of  such  harmonic  currents.  The  reactance  of  the  coil 
for  the  nth  harmonic  is  n  times  as  great  as  for  the  fundamental 
wave;  therefore,  the  higher  harmonics  in  the  current  are  rela- 
tively smaller  than  those  in  the  voltage  wave. 

(c)  Let  now  a  non-sinusoidal  alternating  voltage  be  impressed 
at  the  terminals  of  an  impedance  coil,  and  let  it  be  required  to 
determine  the  wave  form  of  the  current.  The  result  to  be 
expected  will  be  intermediate  between  those  derived  for  a  pure 
resistance  and  a  pure  inductance;  viz.,  the  current  wave  will  be 
more  nearly  of  sine  form  than  the  voltage  wave,  but  not  to  the 
same  extent  as  in  the  case  of  a  pure  inductance. 

Substituting  the  above  given  expansion  for  the  voltage  wave 


CHAP.  VI]        INDUCTANCE,  REACTANCE  AND  IMPEDANCE        71 

into  the  fundamental  eq.  (83),  we  obtain  a  differential  equation  for 
i,  which  equation  some  readers  may  not  be  able  to  solve.  We 
choose,  therefore,  the  opposite  way;  that  is,  we  assume  the  current 
wave  to  be  given,  instead  of  the  voltage  wave,  and  determine  the 
corresponding  voltage  wave  from  eq.  (83).  This  procedure  is 
much  simpler,  because  it  involves  differentiation  instead  of  inte- 
gration. Let  the  current  be  given  in  the  form 

i  =  /i  sin  (u  —  «i)  +  73  sin  3  (u  —  «3)  +  etc.,  where  u  =  2  irft. 

Substituting  this  value  into  eq.  (83)  and  rearranging  the  terms, 

gives 

e  =  [Iir  sin  (u  —  «i)  +  2  irfLIi  cos  (u  —  ai)]  +  [/3r  sin  3  (u  —  a3) 

+  6 Tr/L/g  cos  3  (u  ->3)]  +  etc.  .     .     .     (100) 

This  result  shows  that  each  harmonic  of  the  current  requires  a 
corresponding  harmonic  of  the  voltage,  as  if  it  were  flowing  alone. 
The  total  voltage  is  equal  to  the  sum  of  the  harmonic  voltages. 
Therefore  we  conclude  that,  conversely,  if  the  voltage  were  given, 
the  current  would  be  equal  to  the  sum  of  the  harmonic  currents 
produced  by  the  respective  harmonics  in  the  voltage.  If  the 
impedance  to  the  first  harmonic  is  z\  =  \/r2  +  x2,  that  to  the 
third  harmonic  is  z3  =  Vr2  +  (3  x)2,  and  in  general  the  impedance 
to  the  nth  harmonic  is  zn  =  Vr2  +  (nx)2.  The  phase  displace- 
ment between  the  corresponding  harmonics  of  current  and  voltage 
is  determined  from  the  condition,  tan  <£„  =  nx/r,  or  cos  0n  =  r/zn. 
The  general  conclusion  reached  is  as  follows:  When  the  applied 
voltage  contains  higher  harmonics,  the  total  current  is  found  by 
summing  the  harmonic  currents  due  to  each  harmonic  of  the 
voltage  acting  alone. 

Prob.  1.  The  effective  value  of  the  fundamental  wave  of  an  e.m.f.  is 
110  volts;  it  has  a  pronounced  third  harmonic,  of  24  per  cent  of  the  fun- 
damental wave.  This  voltage  is  applied  across  a  pure  reactance,  equal  to 
5  ohms  for  the  fundamental  frequency.  Calculate  the  current. 

Ans.     22.07  amp. 

Prob.  2.  An  alternating  voltage  is  represented  by  the  expression 
170  sin  250  t  +  62  sin  (1250  t  +  2.3).  It  is  applied  to  an  impedance  coil 
having  an  inductance  of  45  millihenrys  and  a  resistance  of  7  ohms.  Show 
that  the  current  in  amperes  is  equal  to  12.82  sin  (250 1  —  1.015)  + 
1.09  sin  (1250  £  +  0.853). 

24.  The  Extra  or  Transient  Current  in  Opening  and  Clos- 
ing a  Circuit.  Since  an  electric  current  possesses  inertia  in 


72  THE  ELECTRIC  CIRCUIT  [ART.  24 

the  form  of  inductance,  no  current  can  be  established  or  broken 
instantly,  unless  the  applied  electromotive  force  be  infinitely 
large.  Thus,  when  a  large  electromagnet  is  connected  to  a  source 
of  continuous  voltage,  the  current  increases  during  an  appreciable 
interval  before  it  reaches  its  final  value.  Again,  when  the  circuit 
is  broken,  the  current  continues  in  the  form  of  an  arc  through 
the  air  for  an  appreciable  time.  In  a  majority  of  cases  these 
transient  phenomena  at  the  opening  and  closing  of  a  circuit  are  of 
no  practical  importance,  yet  there  are  circumstances  under  which 
they  must  be  taken  into  consideration;  for  instance,  in  switching 
on  and  off  large  amounts  of  energy,  in  high-frequency  oscillations, 
in  highly  inductive  circuits,  etc.  We  shall  consider  here  two 
simple  cases  of  such  extra  currents;  namely,  when  a  circuit  pos- 
sessing resistance  and  inductance  is  connected  to  a  source  of 
(a)  continuous  voltage  and  (b)  sinusoidal  alternating  voltage. 

(a)  Direct  Voltage.  When  e  in  eq.  (83)  is  constant,  one  value 
of  i  which  satisfies  this  equation  is  i  =  e/r,  because  in  this  case 
di/dt  =  0.  However,  this  is  not  the  most  general  solution, 
because  it  is  possible  to  select  an  exponential  expression  in  addi- 
tion to  the  constant  i,  which  will  satisfy  the  equation.  Put 

i  =  «/r  +  CV-'*f (101) 

where  e  is  the  base  of  natural  logarithms,  and  C  and  T  are  certain 
constants.  Substituting  this  value  of  i  into  eq.  (83),  we  get 

Cre-'/'  -  (L/r)Ce-^  =  0, 
or 

T  =  L/r. 

Besides,  i  =  0  when  t  =  0,  so  that  expression  (101)  becomes 
0  =  e/r  +  C,  from  which 

C  =  -e/r, 
and  consequently 

i  =  (e/r)  (1  -  «-*/*) (102) 

In  other  words,  when  a  direct-current  circuit  is  closed,  the  current 
increases  at  first  rapidly,  then  more  and  more  slowly;  and  theo- 
retically it  reaches  its  final  value  of  e/r  only  after  an  infinite  time. 
In  reality,  the  current  becomes  practically  constant  after  a  frac- 
tion of  a  second,  unless  the  inductance  is  exceedingly  large.  The 
factor  T  =  L/r  is  called  the  time  constant  of  the  circuit;  it  deter- 
mines the  rate  of  the  initial  rise  in  current,  and  has  the  dimension 
of  time. 


CHAP.  VI]        INDUCTANCE,  REACTANCE  AND  IMPEDANCE        73 

(b)  Sinusoidal  Voltage.  If  the  voltage  follows  the  law  e  = 
Em  sin  2  irft,  one  solution  of  eq.  (83),  as  we  have  seen  before,  is 
i  =  (Em/z)  sin  (2  irft  —  $),  where  cos<£  =  r/z.  But  this  is  not  the 
most  general  solution,  because  it  is  possible  to  add  to  it  an  expo- 
nential term  of  the  form  Ce~t/T,  and  to  select  the  time  constant  r 
in  such  a  way  that  this  term  will  cancel  in  eq.  (83).  Since  the 
sine  term  of  the  current  alone  satisfies  the  equation,  we  will  find 
as  before  T  =  L/r.  The  constant  C  is  determined  by  the  condi- 
tion that  i  =  0  when  i  =  0,  or 

0  =  -  (Em/z)  sin  0  +  C, 
from  which 

C=  (Em/z}sm4>=.Emx/z\ 

Therefore  the  current 

i  =  (Em/z)  sin  (2*fl  -  <£)  +  (Emx/z2)<--<r/L.     .     .    (103) 

Under  ordinary  conditions  the  exponential  term  becomes  negli- 
gibly small  within  a  fraction  of  a  second,  so  that  it  is  legitimate 
to  consider  the  current  to  be  a  pure  sine-wave,  as  we  have  done 
heretofore.  However,  the  extra  current  may  be  of  importance  in 
transient  phenomena,  for  instance,  at  the  moment  of  closing  a 
circuit. 

The  solutions  (102)  and  (103)  of  eq.  (83)  are  found  above  by 
trials,  because  it  is  assumed  that  the  reader  is  not  familiar  with 
the  general  method  for  the  solution  of  linear  differential  equations  ; 
otherwise,  the  solution  could  have  been  written  directly.  Equa- 
tion (83)  is  of  the  form 

dy/dx  +  Py  =  Q,   ......     (104) 

where  P  and  Q  are  functions  of  x  or  constants.  By  referring  to 
any  book  on  differential  equations,  the  reader  will  find  that  the 
general  solution  of  this  equation  is 


....    (105) 

where 

^  (106) 


,-/P<b 


Prob.  1.  The  current  in  a  coil  due  to  a  constant  e.m.f.  reaches  99  per 
cent  of  its  final  value  within  one  hundredth  of  a  second  after  the  circuit 
is  closed.  Show  that  the  time  constant  of  the  coil  is  equal  to  2.17  milli- 
seconds. 


74  THE  ELECTRIC  CIRCUIT  [ART.  24 

Prob.  2.  Show  that  the  time  constant  may  be  defined  as  the  interval 
of  time  during  which  the  current  reaches  (e  —  l)/e  =  0.632  of  its  final 
value. 

Prob.  3.  Select  the  constants  of  an  alternating-current  circuit  so  as 
to  have  a  power-factor  of  about  80  per  cent;  and  plot  curves  of  (a)  the 
voltage,  (b)  the  sinusoidal  component  of  the  current,  (c)  the  exponential 
component  of  the  current,  and  (d)  the  total  current,  for  the  first  few 
cycles  after  the  circuit  is  closed. 

Prob.  4.  Extend  the  theory  given  above  to  the  case  where  the  cir- 
cuit is  closed  at  an  instant  when  the  alternating  voltage  is  not  equal  to 
zero. 

Prob.  6.   Check  the  solutions  (102)  and  (103),  using  formula  (105). 

Prob.  6.  When  an  impedance,  consisting  of  r  and  L,  is  suddenly  short- 
circuited,  so  that  e  becomes  instantly  equal  to  zero,  show  that  the  line 
current  gradually  disappears  according  to  the  exponential  law  i  =  i()(Trt/L, 
where  i'0  is  the  magnitude  of  the  current  at  the  instant  of  short-circuit. 


CHAPTER  VII 
SUSCEPTANCE  AND  ADMITTANCE 

25.  Concept  of  Susceptance.  The  concept  of  reactance,  as 
introduced  in  Art.  21,  indicates  the  degree  of  difficulty  in  forcing 
an  alternating  current  through  a  coil,  against  the  reaction  of  an 
alternating  magnetic  field.  In  this  respect,  reactance  is  analo- 
gous to  resistance.  We  have  seen,  however,  in  Chapter  I,  that  it 
is  more  convenient  to  use  conductances,  when  resistors  are  con- 
nected in  parallel.  Similarly,  when  reactive  coils  or  reactors  are 
connected  in  parallel,  it  is  more  convenient  in  calculations  to  use 
the  reciprocals  of  their  reactances.  The  reciprocal  of  reactance  is 
called  susceptance,  and  is  usually  denoted  by  the  symbol  6.  Thus, 
by  definition,  the  susceptance 

b=l/x  =  l/(2rfL) (107) 

By  analogy  with  conductance,  one  may  say  that  the  susceptance 
measures  the  degree  of  ease  in  forcing  an  alternating  current 
through  a  coil,  against  the  reaction  of  a  pulsating  magnetic  field. 
Since  reactance  is  measured  in  ohms,  susceptance  is  measured  in 
mhos.  Equation  (87)  becomes 

I  =  bE,. (108) 

it  being  understood  as  before  that  the  current  lags  by  90  degrees 
behind  the  voltage.  The  student  is  reminded  that  the  concept 
of  susceptance,  like  that  of  reactance,  implies  pure  inertia  reaction, 
without  any  ohmic  resistance;  this  limitation  is  very  important 
for  a  clear  understanding  of  the  rest  of  the  chapter. 

When  several  inductive  coils  are  connected  in  parallel,  their 
susceptances  are  simply  added  together,  or 

beq  =  61  +  62  +  etc (109) 

The  proof  is  similar  to  that  for  the  addition  of  conductances  (see 
Art.  3).  Thus,  a  susceptance  of  3  mhos  in  parallel  with  one  of 
2  mhos  gives  a  total  susceptance  of  5  mhos. 

75 


76 


THE  ELECTRIC  CIRCUIT 


[ART.  26 


Prob.  Two  reactive  coils  of  10  and  20  millihenrys  respectively  are 
connected,  first  in  series  and  then  in  parallel,  across  a  40-cycle,  180-volt 
line.  The  ohmic  resistance  of  the  coils  is  negligible.  What  is  the  current 
in  each  case?  Ans.  23.85  amp.;  107.35  amp. 

26.  Concept  of  Admittance.  Let  now  a  pure  inductance  be 
connected  in  parallel  with  a  pure  ohmic  resistance,  across  a  source 
of  alternating  voltage  E  (Fig.  24),  and  let  it  be  required  to  find 


E 


FIG.  24.    A  susceptance  in  parallel 
with  a  conductance. 


FIG.  25.  The  voltage  and  current 
relations  in  the  circuit  shown 
in  Fig.  24. 


the  total  current  through  the  combination.'  The  inductance  can 
be  expressed  as  a  susceptance,  and  the  resistance  as  a  conductance. 
The  current  through  the  susceptance,  according  to  eq.  (108),  is 
bE,  in  quadrature  with  the  voltage  (Fig.  25)  ;  the  current  through 
the  conductance,  according  to  eq.  (2),  is  gE,  in  phase  with  the 
voltage.  The  total  current 

=  E  V^~+&?    .     .     .     (110) 


(Ill) 


and  the  phase  angle  is  determined  from  the  relation 
tan<£  =  Eb/Eg  =  b/g 


cos  0  =  g/Vg*  +  &2.       .....     (112) 

In  the  case  of  a  series  connection,  we  have  found  it  convenient 
to  introduce  the  impedance  z  as  a  symbol  for  Vr2  +  z2.     Simi- 
larly, in  a  parallel  connection  it  is  convenient  to  introduce  the 
abbreviation 

y  =  Vg*  +  b\       ......     (113) 

The  quantity  y  is  called  the  admittance  of  a  circuit,  and  is  measured 
in  mhos,  the  same  as  6  and  g.     Equation  (110)  becomes 

i  =  yE,  .......    (114) 

and  eq.  (112), 

COS0  =  g/y  ........      (115) 


CHAP.  VII]          SUSCEPTANCE  AND  ADMITTANCE 


77 


The  three  quantities  g,  b  and  y  form  a  triangle  (Fig.  25),  in  which 
y  is  the  hypothenuse,  and  the  angle  adjacent  to  g  is  the  phase 
angle  </>.  From  this  triangle  we  obtain  two  useful  relations, 

g  =  y  cos  4> 

b  =  y  sin  0. 

When  there  are  several  susceptances  and  conductances  in  parallel 
(Fig.  26),  the  reactive  and  the  energy  components  of  the  current 
I  ^ 


and 


(116) 


FIG.  26.    Susceptances  and  conductances  in  parallel. 

must  be  added  separately  (Fig.  27).     Therefore,  the  amperes  per 
volt  in  phase  or  the  conductances,  and  the  amperes  per  volt  in 


E&3 


FIG.  27.    The  voltage  and  current  relations  in  the  circuit  shown  in  Fig.  26. 

quadrature  or  the  susceptances,  must  also  be  added  separately,  so 
that  the  equivalent  admittance 

y  =  V(gi  +  02  +  etc.)2  +  (61  +  62  +  etc.)2,     .     (117a) 
and 

tan  c/,  =  (&!  +  62  +  etc.)/(0i  +  g*  +  etc.).    .     .     .     (117b) 

The  student  should  compare  Fig.  27  with  Fig.  23  in  order  to  see 
the  similarity  of  procedure  and  the  difference  in  the  physical 
phenomena  in  the  two  cases.  With  a  series  connection,  it  is  the 
current  that  is  common  to  all  the  parts  of  the  circuit,  while  the 
partial  voltages  are  added  geometrically.  In  a  parallel  combina- 
tion, the  voltage  is  common  to  all  the  branches,  while  the  com- 
ponent currents  are  combined  in  their  proper  phase  relations. 


78 


THE  ELECTRIC   CIRCUIT 


[ART.  27 


The  following  table  gives  the  quantities  defined  in  this  and  the 
preceding  chapter,  in  their  proper  relations. 


Friction 
Effect 

Inertia 
Effect 

Connection 

Result 

Unit 

Resistance  r 

Reactance  x 

Series 

Impedance  z 

Ohm 

Conductance  y 

Susceptance  6 

Parallel 

Admittance  y 

Mho 

-  Prob.  1.  What  susceptance  must  be  connected  in  parallel  with  a 
resistance  of  0.2  ohm,  in  order  to  bring  the  power-factor  of  the  combina- 
tion down  to  80  per  cent?  Also,  what  is  the  value  of  the  resultant  ad- 
mittance? Ans.  3.75  mhos;  6.25  mhos. 

Prob.  2.  Two  electrical  devices  are  connected  in  parallel  to  a  line  of 
voltage  E.  One  device  consumes  a  current  /i  at  a  power-factor  cos  <t>i', 
the  total  line  current  is  7,  lagging  behind  the  voltage  by  an  angle  <£. 
Show  how  to  determine  graphically  the  susceptance  and  the  conductance 
of  both  devices. 

27.   Equivalent  Series  and  Parallel  Combinations.     Let  a 

resistance  ra  be  connected  in  series  with  a  reactance  xa;  also  let 
another  resistance  rp  be  connected  in  parallel  with  a  reactance  xp. 
If  the  values  of  the  resistances  and  reactances  are  so  selected  that 
the  series  combination,  when  connected  to  the  same  source  of 
supply,  will  let  through  the  same  current  at  the  same  power- 
factor  as  the  parallel  combination,  then  the  two  combinations  are 
called  equivalent.  It  is  sometimes  convenient  to  replace  a  given 
series  combination  by  an  equivalent  parallel  combination,  and 
vice  versa.  For  instance,  when  some  parts  of  a  circuit  are  in 
parallel  and  others  in  series,  it  is  convenient  for  numerical  cal- 
culations to  replace  them  all  by  an  equivalent  parallel  or  series 
combination. 

The  problem  is  to  find  the  relation  between  the  four  quanti- 
ties r,,  rp,  x,  and  xp,  if  these  quantities  form  two  equivalent  com- 
binations. According  to  the  above-given  definition,  the  angle  0 
is  the  same  for  both,  and  besides,  according  to  eqs.  (93)  and  (114), 

y-i/«, (us) 

where  y  refers  to  the  parallel  combination  and  z  to  the  equiva- 
lent series  combination.  Combining  now  eqs.  (116),  (95)  and  (96), 
we  have 

lAp  =  9  =  y  cos  0  =  (l/z)  (r./z)  =  r8/z2; 


CHAP.  VII]          SUSCEPTANCE  AND  ADMITTANCE  79 

or 

r.rp  =  z*=  l/</2; (119) 

z.zp  =  z2  =  l/i/2 (120) 

By  means  of  eqs.  (119)  and  (120)  a  series  combination  can  be 
replaced  by  an  equivalent  parallel  combination,  and  vice  versa. 
Instead  of  rp  and  xp,  their  reciprocals,  g  and  b,  may  be  used.  In 
practice,  g  and  &  are  usually  spoken  of  as  the  conductance  and  the 
susceptance  of  either  the  series  or  the  parallel  combination;  but 
it  must  be  clearly  understood  that  they  are  the  reciprocals  of  rp 
and  xp,  and  not  of  r,  and  xt.  If  ra  and  xt  are  given,  it  is  first 
necessary  to  determine  rp  and  xp  from  eqs.  (119)  and  (120),  and 
then  to  take  their  reciprocals.  In  other  words,  for  a  series  circuit 
the  equivalent  conductance  and  susceptance  are 

g  =  r./*> (121) 

and  6  =  xt/zz (122) 

On  the  other  hand,  if  g  and  b  are  given, 

r.  =  g/y2; (123) 

x.  =  b/y* (124) 

The  reciprocals  of  r,  and  x,  are  of  no  practical  importance,  and 
are  not  used  in  this  work. 

Prob.  1.  An  impedance  coil  has  a  reactance  of  7.5  ohms;  the  resist- 
ance of  the  winding  is  2  ohms.  What  are  the  susceptance  and  the  con- 
ductance of  the  equivalent  parallel  combination? 

Ans.     124.3  and  33.2  millimhos. 

Prob.  2.  Check  the  answer  to  the  foregoing  problem  by  actually  cal- 
culating the  current  and  the  power-factor  of  the  series  and  the  parallel 
combinations  at  some  assumed  voltage. 

Prob.  3.  Show  that  rp  and  xp  are  always  larger  than  rt  and  x,  re- 
spectively. Hint:  In  eqs.  (119)  and  (120)  replace  zz  by  ra2  +  x,2,  and 
solve  for  rp  and  xp. 

Prob.  4.  An  apparatus  takes  25  amp.  and  2000  watts  at  110  volts, 
the  current  being  a  lagging  one.  What  are  the  equivalent  conductance 
and  susceptance  of  the  device?  What  are  the  resistance  and  reactance 
in  series  equivalent  to  this  apparatus? 

Ans.    0.165  mho;  0.156  mho;  3.2  ohms;  3.04  ohms. 

Prob.  6.  In  adjusting  a  measuring  instrument,  a  non-inductive  re- 
sistance of  120  ohms  was  used  in  parallel  with  a  choke  coil.  The  imped- 
ance of  the  coil  was  75  ohms,  its  resistance  16  ohms.  In  the  regular 
manufacture  of  the  instrument  it  is  desired  to  use  a  resistance  and  a  re- 
actance in  series.  Determine  their  values,  either  graphically  or  analyti- 
cally. Ans.  ra  =  38.0  ohms;  xa  =  44.3  ohms. 


80  THE  ELECTRIC   CIRCUIT  [ART.  28 

28.  Impedances  in  Parallel  and  Admittances  in  Series.  In 
the  preceding  chapter  we  have  learned  how  to  add  impedances 
in  series,  and  in  this  chapter  how  to  add  admittances  in  parallel. 
Let  now  two  or  more  impedances  be  connected  in  parallel,  and  let 
it  be  required  to  find  the  equivalent  impedance.  This  is  done 
by  replacing  each  of  the  given  impedances  by  an  equivalent 
parallel  combination,  and  then  adding  their  admittances  in  par- 
allel, according  to  the  rule  developed  above.  Conversely,  let  sev- 
eral admittances  be  connected  in  series,  and  let  it  be  required  to 
find  the  equivalent  admittance.  To  solve  this  problem,  each 
parallel  combination  is  replaced  by  an  equivalent  series  combina- 
tion, and  then  the  impedances  are  added  in  series.  The  student 
understands,  of  course,  that  the  addition  in  both  cases  is  geo- 
metric, and  that  only  like  components  can  be  added  algebraically. 
Problems  of  this  kind  occur,  for  instance,  in  the  theory  of  trans- 
mission lines,  transformers,  and  induction  motors;  for  this  reason 
it  is  important  that  the  student  understand  the  equivalent  com- 
binations, and  that  he  acquire  facility  in  changing  from  a  series 
to  a  parallel  combination,  and  vice  versa,  as  is  explained  in  the 
preceding  article. 

Prob.  1.  The  load  of  a  single-phase,  6600-volt  generator  is  estimated 
to  consist  of  1200  kw.  of  lamps,  practically  non-inductive,  and  of  800  kw. 
of  motors,  working  at  an  average  power-factor  of  75  per  cent.  What 
will  be  the  expected  generator  output,  in  amperes,  and  the  power-factor? 
Solution:  The  energy  component  of  the  motor  current  is  800/6.6  =  121.2 
amp.;  the  reactive  component  is  121.2  tan  ^  =  106.8  amp.  The  lamp 
current  is  1200/6.6  =  181.8  amp.  The  total  energy  component  of  the 
generator  current  is  121.2  +  181.8  =  303  amp.  Consequently,  the  total 
generator  current  is  (3032  +  106.S2)*  =  321.3  amp.;  the  power-factor 
is  303/321.3  =  94.3  per  cent. 

Prob.  2.   Check  the  solution  of  the  preceding  problem  graphically. 

Prob.  3.  Three  resistances  of  2,  5  and  10  ohms,  and  two  reactances  of 
4  and  2.5  ohms,  are  all  connected  in  parallel  across  a  250-volt  alternating- 
current  line.  What  are  the  total  current  and  the  power-factor  of  the  com- 
bination? Ans.  258  amp.;  77.5  per  cent. 

Prob.  4.  Three  impedance  coils,  having  ohmic  resistances  of  2,  3  and 
4  ohms  respectively,  and  inductances  of  13,  10  and  22  millihenrys,  are 
connected  in  parallel  across  a  source  of  220-volt,  60-cycle  alternating 
voltage.  Calculate  the  total  current  and  the  power-factor.  Check  the 
solution  graphically.  Ans.  110  amp.;  cos  0  =  0.495. 

Prob.  6.  Solve  the  preceding  problem  for  a  frequency  of  25  cycles  per 
second.  Construct  the  vector  diagrams  of  the  currents  in  both  problems 
to  the  same  scale,  so  as  to  see  the  influence  of  the  frequency. 


CHAP.  VII]  SUSCEPTANCE  AND  ADMITTANCE  81 

Prob.  6.  In  problem  4,  let  the  total  current  be  given  in  magnitude, 
but  not  in  its  phase  position;  assume  the  inductance  of  the  third  coil  to 
be  unknown.  Show  how  to  determine  analytically  and  graphically  the 
vector  of  the  current  in  the  third  coil,  and  the  position  of  the  vector  of 
the  total  current. 

Prob.  7.  The  admittance  of  a  winding  is  0.2  mho;  the  current  through 
the  winding  lags  by  34  degrees  with  respect  to  the  voltage  at  its  terminals. 
Determine  the  resistance  and  the  reactance  of  the  winding. 

Ans.    4.145  ohms;  2.796  ohms. 

Prob.  8.  A  coil  having  a  resistance  of  2.3  ohms  and  a  reactance  of 
5  ohms  is  connected  in  parallel  with  another  coil,  for  which  r  =  3  ohms 
and  x  =  4  ohms.  Calculate  the  resistance  and  the  reactance  of  the 
equivalent  series  circuit.  Ans.  1.36  ohms;  2.255  ohms.  . 

Prob.  9.  The  coils  given  in  the  preceding  problem  are  connected  in 
parallel  across  55  volts.  Calculate  the  total  current,  its  energy  and  re'ac- 
tive  components,  and  the  power-factor  of  the  combination. 

Ans.    20.85  amp.;  10.78  amp.;  17.88  amp.;  cos  <t>  =  0.5165. 


CHAPTER  VIII 
THE  USE  OF  COMPLEX  QUANTITIES 

29.  Addition  and   Subtraction   of   Projections  of  Vectors. 

With  the  explanation  given  in  the  preceding  four  chapters,  the  stu- 
dent is  enabled  to  handle,  by  means  of  vector  diagrams,  problems 
involving  resistances  and  reactances  in  alternating-current  cir- 
cuits. A  number  of  problems  in  transmission-line  calculations 
and  in  the  theory  of  alternating-current  machinery  may  be  solved 
by  the  use  of  such  vector  diagrams.  The  disadvantages  of  the 
graphical  method  are:  (1)  Results  are  usually  obtained  which 
hold  for  one  specific  case  only;  an  analysis  of  the  effect  of  various 
factors  is  often  difficult.  (2)  Some  vectors  may  be  many  times 
smaller  than  others;  for  instance,  the  voltage  drop  in  a  transmis- 
sion line,  as  compared  to  the  line  voltage  itself.  Therefore,  the 
diagram  must  be  drawn  to  a  very  large  scale,  or  else  the  results 
are  not  sufficiently  accurate.  In  addition  to  these  drawbacks, 
some  engineers  object  to  graphical  methods  in  general,  as  involv- 
ing the  use  of  drawing  instruments,  which  may  not  be  convenient. 
On  the  other  hand,  vector  diagrams  are  quite  convenient  in 
some  practical  cases;  moreover,  they  are  helpful  for  the  under- 
standing of  general  relations  in  a  circuit,  without  reference  to 
particular  numerical  values.  Again,  in  some  problems,  the  un- 
known vectors  can  be  calcu- 
lated from  the  vector  diagram 
trigonometrically ,  without  the 
necessity  of  actually  drawing 
it  to  scale. 

It    is    possible    to    treat 
vectors    analytically,     using 
their  projections  on  two  axes, 
FIG.  28.    A  vector  and  its  projections.     as  in  analytic  geometry  (Fig. 

28).     A  vector,  such  as  E, 

can  be  denned  either  by  its  magnitude  and  phase  angle  0,  or 
by   its   projections   e   and    e'    upon    the    axes    of    coordinates. 

82 


CHAP.  VIII]        THE  USE  OF  COMPLEX  QUANTITIES  83 

If  E  and  6  are  given,  the  projections  are  calculated  from  the 
expressions 

e  =  Ecosd',} 

e'  =  Esme.l' 

If  the  projections  are  given,  the  vector  itself  is  determined  in 
magnitude  and  position  from  the  equations 

E  =  (e2  +  e'2)*; (126) 

tan  6  =  e'/e (127) 

In  numerical  computations  it  is  more  convenient  first  to  calculate 
tan  0  from  eq.  (127)  and  then  to  determine  E  from  one  of  the 
eqs.  (125),  using  trigonometric  tables.  This  does  away  with  the 
necessity  for  squaring  the  projections  and  extracting  a  square 
root. 

The  fact  that  e  and  e'  are  components  of  the  vector  E  along 
two  perpendicular  axes  is  expressed  symbolically  thus: 

E  =  e+je' (128) 

Here  j  is  a  symbol  which  indicates  that  the  projection  e'  refers 
to  the  vertical  axis.  This  symbol  must  not  have  any  real  value; 
for  the  time  being,  it  may  be  considered  merely  as  an  abbrevia- 
tion of  the  words  "  along  the  vertical  axis."  The  sign  plus  in 
eq.  (128)  denotes  the  geometric  addition.  The  dot  under  E  sig- 
nifies that  by  E  is  meant  not  only  the  magnitude  of  the  vector, 
but  its  direction  as  well,  the  latter  being  defined  by  the  projec- 
tions. When  the  magnitude  only  is  meant,  the  dot  is  omitted. 

The  foregoing  notation  has  been  introduced  by  Dr.  Charles 
P.  Steinmetz,  and  is  now  universally  used  in  this  country.  Much 
credit  is  also  due  to  Dr.  Steinmetz  for  developing  the  analytic 
method,  used  below,  of  dealing  with  alternating  currents  and 
voltages  by  means  of  their  projections. 

The  addition  and  subtraction  of  vectors  are  reduced  simply  to 
the  addition  and  subtraction  of  projections.  According  to  Fig.  12, 
the  projection  of  a  vector  on  any  axis  is  equal  to  the  sum  of  the 
corresponding  projections  of  its  component  vectors  on  the  same 
axis.  Thus,  if  a  current  is  represented  as  a  vector  by  its  projec- 
tions 50  -f-  j  70  amp.,  and  another  current  by  100  +  j  40  amp., 
the  vector  sum  of  these  currents  is  150  +  j  HO  amp.  Or,  the 
resultant  of  two  voltages,  EI  —  e\  +  jei  and  E2  =  e2  +  jez',  is 
Eeq  =  EI  -\-  Ez  =  (ei  +  e<t)  -\-  j(e\  +  62'). 


84  THE  ELECTRIC  CIRCUIT  [ART.  30 

As  an  illustration,  let  us  solve  problem  3,  Art.  14,  by  the  method 
of  projections.  Take  the  voltage  vector  of  the  first  alternator 
in  the  horizontal  direction,  this  being  the  simplest  assumption. 
This  vector  is  therefore  expressed  as  E\  =  2300  +  j  0.  The  hor- 
izontal projection  of  the  second  vector  is  1800  cos  27°  =  1603.8 
volts,  and  its  vertical  projection  is  1800  sin  27°  =  817.2  volts. 
Both  of  these  projections  are  positive,  because  the  second  vector 
leads  the  first,  and  is  therefore  in  the  first  quadrant.  Thus,  E2  = 
1G03.8  +  j  817.2  volts.  The  resultant  voltage,  Eeq  =  Ev  +  E2  = 
3903.8  +  j  817.2  volts.  For  some  purposes,  it  is  sufficient  to 
leave  the  answer  in  this  form;  if,  however,  the  magnitude  and 
phase  position  are  required,  they  are  found  as  explained  above: 
tan  e  =  817.2/3903.8  =  0.2092;  6  =  11°  49';  cos  9  =  0.9786;  Eeq 
=  3903.8/0.9786  =  3988  volts. 

If  the  terminals  of  the  second  machine  be  reversed,  then  Eeq  = 
EI  —  E2  =  696.2  —  j  817.2.  This  vector  has  a  positive  horizon- 
tal projection  and  a  negative  vertical  projection.  Consequently, 
it  lies  in  the  fourth  quadrant,  and  lags  behind  the  reference  vec- 
tor by  less  than  90  degrees.  Proceeding  as  above,  we  find  6  = 
-49°  32';  Eeq  =  1074  volts. 

Prob.  1.  Solve  problem  1,  Art.  14,  by  the  method  of  projections, 
assuming  the  vector  of  the  first  current  to  be  horizontal. 

Prob.  2.  Check  the  solution  of  problem  4,  Art.  14,  by  the  method  of 
projections. 

v  30.  Rotation  of  Vectors  by  Ninety  Degrees.  In  problems 
involving  reactance,  it  is  necessary  to  multiply  the  vector  of  the 
current  by  the  reactance  of  the  circuit  and  then  turn  it  by  90 
degrees,  in  order  to  determine  the  reactive  drop  in  voltage.  The 
simple  multiplication  of  the  vector  of  current  by  the  reactance 
converts  it  into  a  vector  of  voltage,  and  thus  merely  changes  the 
scale.  But  turning  the  vector  modifies  the  relative  magnitudes 
of  its  projections;  it  is,  therefore,  necessary  to  find  a  relation 
between  the  magnitudes  of  the  original  and  the  new  projections. 
In  the  simplest  case  let  a  vector  E\  be  drawn  along  the  refer- 
ence axis,  or  axis  of  abscissae,  and  let  its  length  be  a.  In  the  sym- 
bolic notation  it  is  represented  as  El  =  a,  the  other  projection 
being  zero.  After  having  been  turned  by  90  degrees  counter-clock- 
wise, the  vector  is  directed  along  the  positive  axis  of  ordinates, 
and  is  symbolically  represented  as  E2  =  ja,  the  horizontal  pro- 


CHAP.  VIII]         THE  USE  OF  COMPLEX  QUANTITIES  85 

jection  being  zero.  Thus,  in  this  particular  case,  a  rotation  by 
90  degrees  is  equivalent  to  a  multiplication  by  j.  . 

It  is  convenient  to  define  j  in  such  a  manner  that  multiplication 
of  any  vector  by  j  will  turn  the  vector  by  90  degrees  in  the  positive 
direction  (counter-clockwise),  while  division  by  j  will  turn  the  vec- 
tor by  90  degrees  in  the  negative  direction  (clockwise).  In  order  to 
find  a  value  of  j  which  satisfies  these  requirements,  let  the  vector 
EZ  be  turned  again  by  90  degrees  counter-clockwise,  being  now 
directed  along  the  negative  X-axis.  Its  expression  is  now  E3  = 
—a.  On  the  other  hand,  the  same  expression  must  be  obtained 
by  multiplying  E2  by  j.  Therefore,  we  have  —  a  =  fa,  or  j2  = 
—  1;  consequently,  j  =  V/— 1.  If  the  original  vector  EI  is  to  be 
turned  by  90  degrees  clockwise,  we  must,  according  to  our  assump- 
tion, divide  it  by  j.  We  then  have  E*  =  a/j,  or,  multiplying  the 
numerator  and  the  denominator  by  j,  #4  =  ja/j2.  If  j2  =  —  1,  as 
it  appears-  to  be  above,  then  Et  =  —ja.  This  checks  with  the 
preceding  result,  because  E^  =  —E*-  It  will  thus  be  seen  that 
the  value  of  j2  =  —  1  satisfies  the  requirements  set  above,  when 
the  original  vector  is  directed  along  one  of  the  axes  of  coordinates. 

Let  now  the  original  vector  EI  (Fig.  29)  have  an  arbitrary 
direction  in  the  first  quadrant,  or  EI  =  a  +  jb.  Multiplying  EI 
by  j  we  must  get  the  vector  Ez,  of  the  v 

same  magnitude,  but  in  the  second 
quadrant  and  perpendicular  to  EI. 
E2  has  a  vertical  projection  equal  to 
the  horizontal  projection  a  of  the 
original  vector  EI;  the  horizontal  pro- 
jection of  E2  is  negative,  and  is  equal 
in  its  absolute  value  to  the  vertical 
projection  6  of  the  vector  E1.  Thus,  FlG'  »•  The  relation  between 
,,  .  ,  the  projections  of  two  vectors, 

the  new  vector  is  expressed  as  pe^endicular  to  each  other. 
E2  =  —  b  +  ja.  On  the  other  hand, 

multiplying  EI  by  j  we  have  jEi  =  ja  +  fb  =  ja  —  b,  which  is 
the  same  as  above.  Therefore,  in  this  case  also  the  assump- 
tion j2  =  —  1  is  correct,  and  leads  to  rotation  by  90  degrees. 
It  is  left  to  the  student  to  verify  the  cases  in  which  the  vector  lies 
in  some  other  quadrant,  and  where  EI  is  divided  by  j,  for  rotation 
by  90  degrees  in  the  negative  direction. 

Expressions  of  the  form  a  +  jb,  where  a  and  6  are  real  quan- 
tities and  j  =  V—  1,  are  called  in  algebra  complex  quantities. 


86  THE  ELECTRIC   CIRCUIT  [ART.  30 

The  student  need  not  be  discouraged  by  the  name,  because  for 
our  purposes  j  is  simply  a  quantity  which  separates  the  two  pro- 
jections of  a  vector,  obeys  the  law  of  multiplication  and  division, 
and  is  of  such  a  nature  that  j2  =-1.  Moreover,  solutions  by 
means  of  complex  quantities  are  quite  as  simple  as  by  other 
methods. 

Prob.  1.  A  current  of  80  +  j  43  amp.  flows  through  a  resistance  of 
2  ohms  in  series  with  a  reactance  of  3  ohms.  Find  the  voltage  drop  across 
the  impedance.  Solution:  The  vector  of  the  voltage  consists  of  two 
components,  representing  the  ohmic  and  the  reactive  drop  respectively. 
The  ohmic  drop,  #,,  is  equal  to  2  (80  +  j  43)  =  160  +  j  86  volts.  To 
find  the  inductive  drop,  EZ,  the  vector  of  the  current  must  be  multiplied 
by  x  =  3,  and  then  turned  by  90  degrees,  in  other  words,  multiplied  by  j. 
Thus,  Ez  =  3j  (80  +J43)  =-129  +  J240  volts.  The  total  voltage 
E  =  Ei  +  Ez  =  31  +  j  326  volts. 

Prob.  2.  Solve  the  preceding  problem  when  the  voltage  is  given  and 
the  current  is  unknown.  First  Solution:  Let  the  unknown  current  be 
represented  by  its  projections  as  i  +  ji'.  We  have,  as  in  the  preceding 
problem, 

2  (i  +  ji')  +  3j  (i  +  ji')  =  31  +  j  326,      .     .     .     (129) 


or,  collecting  the  terms  containing  j, 

(2i-3i')+j(2i'  +  3i)  =  31+J326.     .     .     .     (130) 

This  equation  can  be  satisfied  only  if  the  terms  with  and  without  j 
are  equal  to  each  other  respectively,  because  a  real  quantity  cannot  be 
equal  to  an  imaginary  one.  Or,  from  a  geometric  point  of  view,  the  left- 
hand  side  and  the  right-hand  side  of  eq.  (130)  each  represent  a  vector  by 
its  projections.  But  two  vectors  are  identical  only  when  their  corre- 
sponding projections  are  equal.  Thus,  we  have 

2t-3i'  =  31;  2i'  +  3i  =  326. 

Solving  these  equations  for  i  and  i',  we  find  i  =  80,  i'  =  43,  as  in  the  pre- 
ceding problem.  Second  Solution:  Equation  (129)  can  be  written  in  the 
form  (2  +  3;)  (i  +  ji')  =  31  +  j  326;  or,  i+  ji'=  (31+  j  326)  /  (2  +3j). 
Considering  here  j  as  an  ordinary  algebraic  quantity,  we  can  get  rid  of 
it  in  the  denominator  by  multiplying  both  the  numerator  and  the  denomi- 
nator by  2  -  3  j.  The  result  is 

.      .,      (31  +  j326)(2-3j)      1040  ,    .  559 
~~2»-(3j)«  =l3-+-7l3-' 

or  i  +  ji'  =  80  +  j  43,  as  before. 

Prob.  3.  A  voltage  of  28  +  j  120  volts  applied  to  the  terminals  of  a 
coil  produces  in  it  a  current  equal  to  4  +j  1.5  amp.  Determine  the 
resistance  and  the  reactance  of  the  coil. 

Ans.    r  =  16  ohms;  x  =  24  ohms. 


CHAP.  VIII]         THE  USE  OF  COMPLEX  QUANTITIES  87 

Prob.  4.  Verify  the  answer  to  problem  4,  Art.  28,  by  the  method  of 
projections,  assuming  the  vector  of  the  voltage  to  be  horizontal. 

31.  Impedance  and  Admittance  Expressed  as  Complex  Quan- 
tities or  Operators.  Let  it  be  required  to  find  the  voltage  neces- 
sary to  maintain  a  current  i  +  ji'  through  a  resistance  r  and  react- 
ance x  in  series.  The  voltage  drop  in  the  resistance  is  r  (i  +  ji'} ; 
that  in  the  reactance  is  jx  (i  -i-  ji'}.  Hence,  the  total  voltage  is 
E  =  r(i+ji')+jx(i+ji'),  or 

E  =  e  +  je'  =  (r  +  jx)  (i  +  ji'}.  .  .  .  (131) 
It  is  legitimate  to  factor  out  the  expression  (i  +  ji'),  and  to  treat 
j  as  any  other  algebraic  quantity,  because  j  is  now  assigned  a 
definite  value,  V^l.  Moreover,  eq.  (131)  represents  simply  the 
geometric  addition  of  four  component  vectors,  two  of  them 
directed  along  the  X-axis  and  the  other  two  along  the  F-axis. 
As  long  as  this  interpretation  is  kept  in  mind,  the  terms  may  be 
arranged  in  any  desired  order. 

Equation  (131)  shows  that,  in  order  to  obtain  the  expression 
of  the  voltage  drop  through  an  impedance,  the  current  must  be 
multiplied  by  the  complex  quantity  r  +  jx.  The  expression 
r  -\-jx  is  not  a  vector,  because  it  does  not  stand  for  a  sine-wave, 
but  an  operator  upon  the  vector  of  the  current.  The  operation 
consists,  first,  in  multiplying  the  vector  of  the  current  by  r,  then 
in  multiplying  the  same  vector  by  x  and  turning  it  by  90  degrees 
in  the  positive  direction,  and  finally,  hi  adding  the  two  vectors 
geometrically.  All  these  operations  are  included  in  the  expression 
r  +  jx,  which  is  called  the  impedance  operator. 

In  order  to  get  the  projections  e  and  e'  of  E  from  eq.  (131), 
the  terms  on  the  right-hand  side  must  be  actually  multiplied  and 
the  results  represented  in  the  form  of  a  complex  quantity.  We 
get  then,  separating  the  real  and  the  imaginary  parts, 

E  =  e  +  je'  =  (ri  -  XL')  +  j  (rir  +  xi). 

The  real  and  the  imaginary  parts  on  each  side  of  this  equation  are 
equal  to  each  other  respectively,  because  they  represent  the  pro- 
jections of  the  same  vector  E  upon  the  two  axes.  Consequently 

e  =n-»';l 

e'  =  ri'  +  xi.  j  ' 

In  problems  these  steps  are  best  left  until  the  numerical  values 
have  been  substituted,  in  order  to  avoid  complicated  expressions. 


88  THE  ELECTRIC  CIRCUIT  [AET.  31 

If  the  voltage  and  the  impedance  are  given,  and  it  is  required 
to  find  the  current,  we  get  from  eq.  (131)  the  relation 


In  order  to  reduce  the  right-hand  side  of  this  equation  to  the 
form  of  a  complex  quantity,  we  multiply  the  numerator  and  de- 
nominator by  the  expression  r  —  jx.  This  gives 

,  _  (e+je')(r-jx)  _  re  +  xe'        re'  -  xe 

i+v       -^nr^i         r2  +  X2  +  -V  +  x2 

Equating  the  real  and  the  imaginary  parts  respectively,  we 
obtain 


i'  =  (re'  -  xe)/*.       ' 

Equation  (131)  expresses  the  fact  that  the  voltage  E  is  equal 
to  the  product  of  the  current  by  the  impedance,  if  the  operator 
(r  +  jx)  be  considered  as  the  impedance  of  the  circuit  in  the  com- 
plex notation.     Denote  the  impedance  by  Z,  then 

Z  =  r+jx  ........     (135) 

Here  capital  Z  is  used  to  indicate  that  it  is  a  complex  quantity, 
as  distinguished  from  the  numerical  value  z  of  the  same  imped- 
ance. The  letter  is  not  provided  with  a  dot,  because  Z  is  not  a 
vector,  but  an  operator. 

In  the  abbreviated  notation,  eq.  (131)  becomes 

E  =  IZ  .........     (136)  • 

In  this  expression  each  letter  stands  for  a  complex  quantity,  so 
that  when  actual  numerical  or  algebraic  relations  are  necessary, 
the  expression  must  again  be  expanded  into  (131)  and  the  multi- 
plication of  the  two  complex  quantities  actually  performed. 

Instead  of  dividing  the  voltage  by  the  operator  (r  +  jx)  and 
then  eliminating  j  from  the  denominator,  it  is  more  convenient 
to  introduce  another  operator  by  which  the  voltage  must  be  multi- 
plied in  order  to  obtain  the  current.  It  will  be  remembered  from 
Art.  26,  that  a  voltage  must  be  multiplied  by  an  admittance  in 
order  to  get  the  current.  Consequently  the  operator  in  question 
must  be  expected  to  have  the  elements  and  the  dimensions  of  an 
admittance.  Replacing  the  given  series  combination  by  an  equiv- 
alent parallel  combination  (Art.  27),  the  unknown  current  is  split 
into  a  component  Eg  through  a  pure  conductance,  and  a  compo- 
nent —  jEb  through  a  pure  susceptance  in  parallel  with  the  con- 


CHAP.  VIII]        THE  USE  OF  COMPLEX  QUANTITIES  89 

ductance.  The  latter  component  is  provided  with  the  prefix  —j, 
because  it  lags  by  90  degrees  behind  the  voltage.  Thus,  the  total 
current 

!=E(g-jb)  .......     (137) 

The  expression 

Y  =  g-jb  ........     (138) 

is  called  the  admittance  operator.  The  symbol  Y,  like  the  symbol 
Z  above,  is  not  provided  with  a  dot  because  it  is  not  a  vector. 
Combining  the  two  preceding  equations  gives 

I  =  YE  ........     (139) 

Equations  (136)  and  (139)  represent  generalized  forms  of  Ohm's 
law  for  alternating  currents,  corresponding  to  the  simple  expres- 
sions (1)  and  (2)  for  direct  current.  Equation  (139)  is  an  abbre- 
viated form  of  the  relation 

i+ji'  =  (e  +  je')(g-jb)  .....     (140) 

Multiplying  out  and  equating  the  real  and  the  imaginary  parts 
on  both  sides  of  this  equation,  we  get 


The  relations  between  r,  x  and  z  on  one  hand,  and  g,  b  and  y  on 
the  other,  are  deduced  in  Art.  27;  it  being  understood,  of  course, 
that  in  the  present  treatment  the  two  combinations  are  equivalent. 
Equations  (136)  and  (139)  imply  that 

YZ  =  1,      .......     (142) 

(r+jx)(g-jb)  =  1. 

Substituting  into  this  last  equation  the  values  of  g  and  6  from 
eqs.  (121)  and  (122)  and  performing  the  multiplication,  it  will  be 
found  that  the  equation  is  reduced  to  the  identity  1  =  1,  this 
being  a  check  on  eq.  (142). 

With  the  abbreviated  notation  of  complex  quantities,  using  the 
symbols^,  /,  Z  and  F,  alternating-current  problems  are  solved 
almost  as  easily  as  direct-current  problems.  Either  the  impedance 
operator  or  the  admittance  operator  is  used,  depending  upon  the 
relative  connection  of  the  parts  of  the  circuit,  whether  parallel 
or  series.  In  many  cases  the  abbreviated  notation  may  be  pre- 
served until  the  solution  has  been  obtained,  the  projections  of 
the  vectors,  e  +  je'  and  i  +  ji',  and  the  expanded  forms  of  the 


90  THE  ELECTRIC  CIRCUIT  [ART.  31 

operators,  r  +  jx  and  g  -  jb,  only  then  being  substituted  in  a 
numerical  form  to  get  the  final  answer. 

Prob.  1.  Find,  by  means  of  complex  quantities,  the  voltage  required 
in  problem  5,  Art.  21.  Solution:  Assume  the  vector  of  the  current  to 
be  the  reference  vector.  At  the  power  house  cos  <f>  =  7520/(66^X  147)  = 
0.775;  sin</>  =  0.632.  The  generator  voltage  is  66  X  0.775  +j  66  X 
0.632  =  51.15  +.7  41.71  kilovolts.  The  voltage  drop  in  the  line  is 
147(45  +j  83)  =  6615  +j  12,200  volts.  Hence,  the  load  voltage  is 
(51.15  -  6.61)  +  ./(41.71  -  12.2)  =  44.54  +j  29.51  kilovolts.  The  nu- 
merical value  of  the  load  voltage  is  (44.S42  +  29 .512)*  =  53.43  kilo- 
volts. 

Prob.  2.  Determine  analytically  the  resistance  r2  required  in  problem 
2,  Art.  22. 

Prob.  3.  A  voltage  equal  to  180  +  j  75  produces  a  current  of  7  + 
j  1.5  amp.  What  is  the  impedance  of  the  circuit? 

Ans.    26.78  +  j  4.97  ohms. 

Prob.  4.  Power  is  transmitted  from  a  single-phase  alternator  to  a 
load  consisting  of  a  resistance  of  1.17  ohms  in  series  with  a  reactance  of 
0.67  ohm.  The  generator  voltage  is  2300,  and  the  impedance  of  the 
transmission  line  is  0.085  -f  j  0.013  ohm.  Determine  (a)  the  line  cur- 
rent; (b)  the  voltage  drop  in  the  line;  (c)  the  receiver  voltage.  Take 
the  generator  voltage  as  the  reference  vector. 

Ans.  (a)  1413.6  -j  769.6  amp.;  (b)  130.1  -j  47  volts;  (c)  2169.9  + 
j  47  volts.  Use  the  admittance  operator  to  obtain  the  current,  and  the 
impedance  operator  to  calculate  the  line  drop. 

Prob.  5.  A  voltage,  e  +  je',  is  impressed  across  the  impedances 
r\  +  jxi  and  r2  +  jxz  in  parallel.  Find  the  total  current.  Solution: 
The  total  conductance  is  g  =  n/Zi2  +  r2/222,  and  the  total  susceptance  is 
b  =  Ji/zi2  +  z2/z22.  Hence,  the  current  i  +  ji'  =  (e  +  je')  (g  —  jb)  = 
(eg  +  e'b)  +  j  (e'g  -  cb). 

Prob.  6.  Extend  the  solution  of  the  preceding  problem  to  the  case 
in  which  more  than  two  impedances  are  in  parallel. 

Ans.     i  +  jV  =  [e  s(r/«»)  +  e'z(z/*)]  +  j  [e'z(r/s»)  -  e  S(z/z2)]. 

Prob.  7.  Two  impedances,  r\  +  jx\  and  r2  +  jx2,  in  parallel,  are 
connected  in  series  with  a  third  impedance  r  +  jx.  Show  how  to  deter- 
mine the  total  voltage,  knowing  the  total  current  i  +  ji' ;  or,  how  to  find 
the  expression  for  the  total  current  when  the  total  voltage  e  +  je'  is 
given. 

Prob.  8.  Show  how  to  solve  the  preceding  problem  when  both  the 
current  and  the  voltage  are  given,  but  either  the  impedance  n  +  jxi  or 
the  impedance  r  +  jx  is  unknown. 


CHAPTER  IX 
THE  USE  OF  COMPLEX  QUANTITIES  —  (Continued) 

32.  Power  and  Phase  Displacement  Expressed  by  Projec- 
tions of  Vectors.  Let  an  alternator  supply  a  current  /  =  i  -f-  ji' 
at  a  voltage  E  =  e  -f  je',  and  let  it  be  required  to  calculate  the 
power  output  of  the  generator.  The  expression  for  the  average 
power  is  P  =  El  cos  <£,  where  <£  is  the  phase  displacement  between 
E  and  /.  The  angle  <£  is  the  difference  between  the  angles  de  and 
0»  which  the  vectors  E  and  /  respectively  form  with  the  reference 
axis.  Hence,  we  have 

P  =  El  cos  <f>  =  El  cos  (0.  -  6i) 

=  E  cos  de  •  I  cos  0i  +  E  sin  de  •  /  sin  0». 

Remembering  that  E  cos  de,  Esin6e,  etc.,  represent  the  projec- 
tions of  the  given  vectors  on  the  axes  of  coordinates,  we  have 
simply 

P  =  ei  +  e'i' (143) 

Another  way  of  deducing  expression  (143)  is  to  resolve  the  given 
vectors  of  current  and  voltage  into  their  components  along  the 
axes  of  coordinates,  and  to  consider  the  contribution  of  each  pro- 
jection to  the  total  power.  The  projections  e  and  i,  being  in 
phase,  give  the  power  ei.  Similarly,  the  projections  e'  and  i'  give 
the  power  e'i'.  The  projection  i'  of  the  current  gives  zero  average 
power  with  the  projection  e  of  the  voltage,  the  two  being  in  phase 
quadrature.  For  the  same  reason  the  average  power  resulting 
from  e'  and  i  is  equal  to  zero.  Thus,  ei  +  e'i'  represents  the 
total  average  power. 

To  find  the  phase  displacement,  or  the  power-factor  of  the 
output,  we  write 

„  N        tan  6e  —  tan  0j 

tan  0  =  tan  (0,  -  00  =  ^—.  — ^- , 

1  +  tan  0e  tan  0< 

....... 

Knowing  tan  </>,  its  cosine  is  found  from  trigonometric  tables.1 

1  Or  else  the  power-factor,  cos  <f>  =  cos  (Oe  —  Oi),  can  be  found  from  the 
relations  Oe  =  tan-1  e'/e,  and  ^  =  tan-1  i' /i. 

91 


92  THE  ELECTRIC  CIRCUIT  [ART.  32 

Power-factor  can  also  be  determined  directly  from  the  expres- 
sion 

cos 0  =  P/EI  =  (ei  +  eY)/[(e2  +  e'2)  (t2  +  i'2)]*,       (145) 

but  the  calculations  are  more  involved  than  when  formula  (144) 
is  used. 

The  power  calculated  by  means  of  formula  (143)  sometimes 
comes  out  negative,  if  some  of  the  projections  of  E  and  /  are 
negative.  The  interpretation  is  that  the  phase  displacement 
between  the  current  and  the  voltage  is  over  90  degrees,  so  that 
power  is  being  supplied  to  the  machine,  instead  of  being  delivered 
by  it.  In  other  words,  the  machine  acts  as  a  motor  and  not  as  a 
generator.  Tan  <j>  in  formula  (144)  may  also  be  negative,  which 
means  either  that  the  current  is  leading,  or  that  it  is  lagging  by 
an  angle  larger  than  90  degrees.  The  question  is  decided  by 
reference  to  the  sign  of  the  power. 

For  the  reactive  power  (Art.  19)  we  have 
Pr  =  EI  sin  0  =  EI  sin  (0e  -  0{)  =  E  sin  del  cos  0t-  -  E  cos  eel  sin  0,, 

or  PT  =  e'i  -  ei' (146) 

The  apparent  power  is 

Pa  =  (#  +  i'rf  (e2  +  e'2)* (147) 

However,  it  is  sometimes  more  convenient  to  determine  the  appar- 
ent power  from  the  relation 

Pa  =  P/  cos  0, (148) 

where  P  is  calculated  from  eq.  (143),  and  cos  0  is  found  from 
trigonometric  tables,  knowing  tan  0  from  eq.  (144). 

Prob.  1.  The  terminal  voltage  of  an  alternator  is  5370  +.7  735;  the 
line  current  is  173  -  j  47  amp.  Calculate  the  output  of  the  machine  and 
the  power-factor  of  the  load.  Ans.  894.5  kilowatts;  92  per  cent. 

Prob.  2.  In  the  preceding  problem,  what  must  be  the  projection  of  the 
current  upon  the  7-axis  in  order  that  the  power  shall  become  zero? 

Ans.    i'=  -1264  amp. 

Prob.  3.  Let  the  line  current  in  problem  1  be  —  58+jl2  amp. 
Explain  the  negative  sign  of  the  power  and  the  plus  sign  of  tan  <f>.  Draw 
the  vectors  of  the  current  and  voltage. 

Prob.  4.  A  synchronous  machine  generates  a  voltage  equal  to  2300  - 
j  50  volts,  and  supplies  a  current,  through  an  impedance  of  5  +  j  50 
ohms,  to  another  synchronous  machine  generating  a  counter-e.m.f.  of 
2300  +  j  50  volts.  What  is  the  power  output  of  the  first  machine?  Is 
the  current  leading  or  lagging?  Make  clear  to  yourself  the  physical 
meaning  of  the  answer.  Ans.  —4.55  kw.;  <t>  =  173°  3'  lagging. 


CHAP.  IX]  THE  USE  OF  COMPLEX  QUANTITIES  93 

Prob.  6.  A  current  of  350  —  j  75  amp.  is  maintained  through  an  im- 
pedance, the  power  output  being  952  kw.  at  a  power-factor  of  86  per  cent 
lagging.  Find  the  voltage  across  the  impedance.  Hint:  Solve  eqs.  (143) 
and  (144)  together  for  the  unknown  projections  e  and  e'. 

Ans.    2930  +  j  987  volts. 

Prob.  6.  Solve  problem  5  by  calculating  the  value  of  the  impedance, 
and  multiplying  the  impedanee  by  the  current.  Hint:  power  =  I-r; 
x  =  r  tan  <f>. 

Prob.  7.   Solve  problem  5,  using  the  expression  for  the  reactive  power. 

^/ 

33.  Vectors  and  Operators  in  Polar  Coordinates.  Instead  of 
representing  a  vector  by  its  orthogonal  projections,  as  in  eq.  (128), 
it  is  sometimes  more  convenient  to  express  the  same  vector  as  a 
complex  quantity  in  terms  of  its  magnitude  and  direction.  Sub- 
stituting the  values  of  e  and  e'  from  eqs.  (125)  into  eq.  (128),  we 
obtain 

#  =#(cos0+./sin0) (149) 

Similarly,  a  current  in  phase  with  this  voltage  is  expressed  as 

7  =  7(cos0+.7'sin0), (150) 

while  a  current  lagging  by  an  angle  0  behind  the  voltage  E  is 
represented  by  the  equation 

1=1  [cos  (0  -  0)  +  j  sin  (0  -  0)].     .     .     .     (151) 

When  the  vectors  of  currents  and  voltages  are  expressed  in  the 
trigonometric  form  shown  above,  it  is  convenient  to  use  the 
operators  Z  and  Y  in  a  similar  form.  Substituting  the  values  of 
r  and  x  from  eqs.  (95)  and  (96)  into  eq.  (135),  we  get 

Z  =  z  (cos  0  +  j  sin  0) (152) 

In  a  similar  manner,  using  eqs.  (116)  in  eq.  (138),  gives 

Y  =  y(cos0  -jsintfO (153) 

When  calculating  the  voltage  drop  IZ  or  the  current  E/Z  it 
is  necessary  to  find  the  product  or  the  ratio  of  two  complex 
expressions  of  the  form  cos  0  +  j  sin  0.  By  actually  performing 
the  multiplication  and  separating  the  real  from  the  imaginary 
term  we  find  that 

(cos  0+  j  sin  0)  (cos  0+  j  sin  0)  =  cos  (0+0)  +  j  sin  (0+0).    (154) 

This  gives  a  simple  rule  for  the  multiplication  of  two  or  more 
complex  quantities  in  the  trigonometric  form.  In  order  to  deduce 


94  THE  ELECTRIC  CIRCUIT  [ART.  33 

a  similar  rule  for  division,  we  observe  that 


cos<>  —  jsm<f>  =  cos  (  —  0)  +  jsm(  —  <£).     .     .     (155) 

This  relation  is  easily  verified  by  multiplying  the  numerator  and 
the  denominator  of  the  left-hand  side  of  the  equation  by  cos  <j>  - 
j  sin  <{>,  so  as  to  get  rid  of  the  complex  quantity  in  the  denomi- 
nator. Equation  (155)  leads  to  the  following  rule  for  the  division 
of  complex  quantities  in  trigonometric  form  : 

(0-«/)).    (156) 


Thus,  for  instance,  if  the  current  given  by  eq.  (150)  flows  through 
an  impedance  expressed  by  eq.  (152),  the  required  terminal  volt- 

age is 

E  =  IZ  =  Iz  [cos  (0  +  0)  +  j  sin  (0  +  </>)],       .     (157) 

which  result  simply  means  that  the  voltage  is  equal  to  Iz  and 
leads  the  current  by  the  angle  0. 

The  operator  given  by  eq.  (152)  multiplies  a  vector  by  z  and 
turns  it  by  the  angle  <£  in  the  positive  direction.  Hence,  the 
operator  (cos  <j>  +  j  sin  0)  simply  turns  a  vector  by  the  angle  <£, 
without  changing  its  length.  Thus,  if  it  be  required  to  turn  a 
vector  A  =  a  +  jar  by  an  angle  a  in  the  positive  direction,  the 
projections  of  the  new  vector  are  found  from  the  following  expres- 
sion: 

(a  +  ja'}  (cos  a  +  j  sin  a)  =  (a  cos  a  —  a'  sin  a) 

+  j  (a'  cos  a  +  a  sin  a)  .....     (158) 

Of  course,  the  same  result  could  be  obtained  by  first  calculating 
the  angle  6  which  the  vector  A  forms  with  the  reference  axis,  from 
the  relation  tan  6  =  a'  /a,  and  then  determining  the  new  projec- 
tions A  cos  (6  +  a)  and  A  sin  (6  +  a). 

Voltage  Regulation  of  a  Transmission  Line.1  As  an  example  of 
the  use  of  complex  quantities  in  the  trigonometric  form,  let  us 
consider  the  voltage  regulation  of  a  single-phase  transmission 
line.  Let  the  resistance  and  reactance  of  the  line,  and  the  gen- 
erator voltage  EI,  be  given;  and  let  it  be  required  to  determine 
the  receiver  voltage  E2  for  a  given  current  I  and  a  given  power- 

1  The  electrostatic  capacity  of  the  line  is  disregarded  here;  a  complete 
treatment  of  the  regulation  of  a  transmission  line,  taking  into  account  the 
capacity  and  leakage,  is  given  in  Arts.  68  and  69  at  the  end  of  the  book. 


CHAP.  IX]  THE  USE  OF  COMPLEX  QUANTITIES  95 

factor  of  the  load  cos  0'.     In  the  symbolic  notation  we  have 

E^Ei  +  IZ, (159) 

where  the  impedance  Z  of  the  line  is  known,  and  is  expressed  by 
eq.  (152).  The  phase  angle  0  refers  to  the  line,  the  angle  0'  to 
the  load. 

When  actually  solving  an  equation  such  as  (159),  it  is  highly 
important  to  select  the  reference  axis  in  the  most  advantageous 
way,  so  as  to  simplify  the  calculations  as  much  as  possible.  In 
the  case  under  consideration,  it  is  convenient  to  select  the  refer- 
ence axis  in  the  direction  of  E2,  because  then  E2  is  determined 
by  its  magnitude  alone,  the  direction  angle  being  equal  to  zero. 
The  generator  voltage  is  expressed  by  EI  (cos  6  +  j  sin  0),  where 
the  magnitude  of  E\  is  given,  but  the  angle  6  is  unknown.  The 
current  lags  by  the  angle  0'  behind  E2,  and  therefore  is  expressed 
by  the  formula  /  (cos  0'  —  j  sin  00-  Thus  eq.  (159)  becomes 

E!  (cos  6  +  j  sin  0)  =  E2  +  Iz  [cos  (0  -  00  +  j  sin  (0  -  0')]-  (160) 
Equating  the  real  and  the  imaginary  parts  gives 

#1COS0  =  #2  +  /ZCOS(0-00;        •       •      -       (161) 

EI  sin  6  =  Iz  sin  (0  -  0') (162) 

From  eq.  (162) 

sin  e  =  (Iz/EJ  sin  (0  -  0') (163) 

Knowing  9,  we  find  from  eq.  (161) 

E2  =  #1  cos  0-  Iz  cos  (0-0').     .     .     .     (164) 

In  practice,  one  is  usually  required  to  determine  the  voltage 
regulation  of  the  line.  According  to  the  definition  adopted  by 
the  American  Institute  of  Electrical  Engineers  (Standardization 
Rules,  Art.  187), 

per  cent  regulation  =  100  (E2  -  Ez)/E2,    .     .     (165) 

where  E2  is  the  value  of  E2  at  no  load.  But  here  E2  =  EI,  be- 
cause the  electrostatic  capacity  of  the  line  is  neglected.  It  is  pos- 
sible to  determine  the  difference  EI  —  E2  directly  from  eq.  (161), 
by  substituting  for  cos  0  the  expression  1  —  2  sin2  £  0.  We  obtain 
then 

A#  =  EI  -  E2  =  Iz  eos  (0  -0')+2  E!  sin2  \  6.        (166) 

Equation  (165)  becomes 

p«r  cent  regulation  =  100  kEKEi  -  A#).       .     (167) 


96 


THE  ELECTRIC  CIRCUIT 


[ART.  33 


When  it  is  required  to  calculate  the  voltage  regulation  for  several 
loads,  the  computations  are  conveniently  arranged  in  a  table 
of  the  following  form: 


I...-L.I 

current 

/ 

Line 
drop 

u 

Load 
power- 
factor 
coa<t>' 

Angle 

«-*' 

sin  <*-*') 

C08(0-<£') 

sin. 

Angle 

e 

sin?  i  6 

A£ 

Per  cent 
regula- 
tion 

In  practice,  the  voltage  regulation  is  usually  required  for  a 
certain  load  of  P2  watts,  so  that,  strictly  speaking,  the  current  / 
is  not  known.  But,  since  E2  is  not  much  different  from  Ei,  it 
is  an  easy  matter  to  estimate  the  current  with  sufficient  accuracy. 
Or  else  a  curve  of  voltage  regulation  is  plotted  against  the  load 
as  abscissae,  so  that  the  regulation  may  be  read  off  at  any  desired 
load.  It  is  possible  to  solve  the  problem  exactly,  by  using  for 
/  its  expression  P2/(E2  cos  <£')  in  eqs.  (161)  and  (162).  In  this 
case  the  equations  are  squared  and  added,  so  as  to  eliminate  6. 
This  gives  a  biquadratic  equation  for  Ez,  from  which  the  receiver 
voltage  can  be  computed. 

This  problem  can  also  be  solved  when  the  complex  quantities 
are  expressed  in  the  orthogonal  form,  instead  of  the  trigonometric 
form  here  used.  The  student  is  urged  to  work  out  the  details,  in 
order  to  become  thoroughly  familiar  with  complex  quantities  in 
both  forms. 

Prob.  1.  A  vector  72  +  j  53  must  be  turned  by  25  degrees  in  the  nega- 
tive direction.  What  are  its  new  projections?  Ans.  87.65  +  j  17.6. 

Prob.  2.  A  single-phase  aluminum  line  is  to  be  built  from  a  power 
house,  at  which  a  voltage  of  11,500  is  maintained  at  a  frequency  of  50 
cycles,  to  a  point  25  km.  distant.  When  a  current  of  60  amp.  at  80  per 
cent  lagging  power-factor  is  delivered  at  the  receiver  end,  the  power  loss 
in  the  line  must  not  exceed  10  per  cent  of  the  useful  power.  What  must 
be  the  size  of  the  conductor,  and  what  will  be  the  per  cent  voltage  regula- 
tion at  this  load?  The  spacing  between  the  wires  is  to  be  61  cm. 

Ans.     No.  0000  B.  &  S.;   11.4  per  cent. 

Prob.  3.  Check  the  answer  to  the  preceding  problem  graphically. 


CHAP.  IX]  THE  USE  OF  COMPLEX  QUANTITIES  97 

Prob.  4.  Explain  the  theory  of  Mershon's  diagram  found  in  various 
electrical  handbooks  and  pocketbooks,  and  check  by  means  of  it  the 
answer  to  problem  2. 

Prob.  6.  Show  how  to  determine  the  voltage  regulation  of  a  trans- 
mission line  when  the  receiver  voltage  is  given. 

Prob.  6.  Show  how  to  calculate  the  receiver  voltage  E2  from  eq.  (159), 
using  the  orthogonal  projections  of  the  vectors  and  operators.  Discuss 
the  relative  advantages  and  disadvantages  of  the  rectangular  and  polar 
coordinates  in  this  case. 

34.  Vectors  and  Operators  Expressed  as  Exponential  Func- 

tions.1 Expressions  (149)  to  (153)  are  sometimes  written  in  the 
exponential  form,  using  the  identity 

cos  e  +  j  sin  6  =  e#,       .....     (168) 

where  e  is  the  base  of  natural  logarithms.  This  important  equa- 
tion follows  from  the  well-known  expansions  for  sin  9,  cos  0  and 
e9,  obtained  by  Maclaurin's  Theorem  in  calculus;  namely, 


The  last  series,  when  j  d  is  substituted  for  9,  becomes 


Substituting  these  values  into  eq.  (168),  it  is  found  to  be  an  iden- 
tity.    Thus,  we  have 

E  =  E  (cos  6+j  sin  9)  =  E&  .....  (169) 
Similarly,  the  impedance  operator  becomes 

Z  =  z  (cos  </>  +  j  sin  0)  =  ze1'*,  ....  (170) 
and  the  admittance  operator 

Y  =  y  (cos  0  -  j  sin  <£)  =  y€-*+.  .  .  .  (171) 
If,  for  instance,  a  current  is  given  as  /  =  Ie}0  and  if  it  flows  through 
an  impedance  Z  =  Zf?*,  the  required  voltage  is  found  by  multi- 
plying these  two  expressions,  or 

E  =  IZ  =  IzeW+<»  ......     (172) 

1  This  article  may  be  omitted  if  desired,  without  impairing  the  conti- 
nuity of  the  treatment  in  the  rest  of  the  book. 


<)8  THE  ELECTRIC  CIRCUIT  [ART.  34 

This  shows  that  the  absolute  value  of  the  voltage  is  Iz  and  that 
it  leads  the  current  by  the  angle  0.  Equation  (172)  corresponds 
to  eq.  (157)  in  trigonometric  notation.  The  projections  of  the 
current  vector  are  /  cos  0  and  /  sin  6;  while  the  projections  of 
the  voltage  vector  are  Iz  cos  (6  +  </>)  and  Iz  sin  (6  +  </>).  Thus, 
it  is  always  possible  to  change  from  the  exponential  form  to  the 
trigonometric  form,  and  finally  to  the  orthogonal  projections,  or 
vice  versa.  The  exponential  form  is  more  concise,  and  possesses 
marked  advantages  in  the  solution  of  some  advanced  problems 
relating  to  alternating  currents  and  oscillations.1  However,  for 
the  simple  problems  treated  in  this  book,  the  plain  algebraic  nota- 
tion a  +  ja'  and  the  trigonometric  notation  A  (cos  a  +  j  sin  a) 
are  amply  sufficient.  It  is  deemed  advisable  to  explain  the  expo- 
nential notation  here  in  order  to  enable  the  student  to  read  books 
and  magazine  articles  in  which  it  is  employed. 

1  See  for  instance  J.  J.  Thomson,  Recent  Researches  in  Electricity  and 
Magnetism. 


CHAPTER  X 
POLYPHASE  SYSTEMS 

35.  Two-phase  System.  The  student  knows  from  his  ele- 
mentary work  that  the  induction  motor  operates  on  the  principle 
of  the  revolving  magnetic  field,  and  that  such  a  field  is  produced 
by  a  combination  of  two  or  more  alternating  currents  differing 
in  phase.  An  electric  circuit  upon  which  are  impressed  two  or 
more  waves  of  e.m.f.  having  definite  phase  displacements  is  called 
a  polyphase  system.  A  large  majority  of  the  alternating-current 
circuits  used  in  practice  hi  the  generation  and  transmission  of 
electrical  energy  are  polyphase  systems;  it  is  therefore  essential 
that  the  student  become  familiar  with  the  current  and  voltage 
relations  in  such  circuits. 

Theoretically,  the  simplest  polyphase  system  is  a  four-wire 
two-phase  system  (Fig.  30),  although  it  is  not  the  most  econom- 


•0 
0 

0 
0 

Phase  1 

N 

g 

o 

-c 

0 

Generator 

Receiver 

Phase  2          > 

FIG.  30.    A  four-wire  two-phase  system  with  two  independent  circuits. 

ical  one  hi  practice.  The  two  generator  windings  are  independ- 
ent, and  are  relatively  displaced  by  ninety  electrical  degrees. 
The  two  alternating  voltages  induced  in  these  windings  are  there- 
fore displaced  in  time  phase  by  a  quarter  of  a  cycle.  Each  phase 
may  be  used  separately,  for  instance  for  lighting,  or  both  phases 
may  be  combined  in  the  windings  of  a  synchronous  or  induction 
motor,  for  the  production  of  a  revolving  magnetic  field.  Each 
phase  can  be  treated  separately,  as  if  it  belonged  to  an  independ- 
ent single-phase  system. 


100 


THE  ELECTRIC  CIRCUIT 


[AST.  35 


Some  economy  in  line  conductors  and  insulators  is  achieved  by 
combining  two  conductors  belonging  to  different  phases  into  one 
return  conductor  (Fig.  31).  Such  a  system  is  called  a  three-wire 


o  Generator 


Fid.  31.    A  three-wire  two-phase  system. 

two-phase  system.     The  current  and  voltage  relations  for  a  balanced 
load  and  a  lagging  current  are  shown  in  Fig.  32.     The  vectors  EI 

and  EZ  represent  the  voltages 
induced  in  the  two  generator 
or  transformer  windings  from 
the  point  0  to  the  points  A 
and  B  respectively;  in  other 
words,  they  are  the  voltages 
between  each  phase  wire  and 
the  return  wire.  The  vector 
Eiz*  is  the  geometric  difference 
of  the  two,  and  represents  the 
voltage  between  the  two  phase 
wires.  That  £"12  is  the  differ- 
ence and  not  the  sum  of  EI 
Fiu.  32.  A  vector  diagram  of  currents  in-  •>  \  ,\ 

and  voltages  for  the  two-phase  system    and     E*     1S     PrOVed     ^     the 
shown  in  Fig.  31.  following   reasoning:    Let   the 

wire  OGG'O'  be   permanently 

grounded,  so  that  its  potential  is  zero.  Let  the  potential  of  the 
wire  A  A'  at  a  certain  instant  be  for  example  100  volts  above 
the  ground,  and  that  of  the  wire  BB'  60  volts  above  the  ground. 
Then  the  difference  of  potential,  or  the  voltage  between  A  A'  and 
BB',  is  40  volts.  The  same  reasoning  applies  to  every  instant, 
so  that  the  vector  of  the  voltage  between  A  A'  and  BB'  is  the 
geometric  difference  between  EI  and  E*,  which  are  the  vectors 
of  the  voltages  between  the  points  A  and  0,  and  B  and  0  re- 
*  Pronounced  E—  one  —  two,  and  not  E  sub  twelve. 


CHAP.  X] 


POLYPHASE  SYSTEMS 


101 


spectively.  That  is,  the  voltage  between  A  and  B  is  represented 
in  phase  and  magnitude  by  the  vector  connecting  the  ends  of  the 
vectors  EI  and  Ez.  Numerically, 

EIZ  =  E1V2  =  E2V2 (173) 

The  currents  in  the  conductors  A  A'  and  BE'  are  represented 
by  the  vectors  l\  and  72,  lagging  by  an  angle  </>  with  respect  to  the 
corresponding  voltages.  The  current  in  the  return  conductor  is 
the  geometric  sum  of  the  two  phase  currents,  and  is  represented 
by  the  diagonal  vector  /i2.  It  will  thus  be  seen  that  the  common 
return  current  is  A/2  times  as  large  as  each  component  current, 
or 

In  =  Ii  V2  =  72  V2 (174) 

If  it  is  desired  to  have  the  same  current  density  hi  each  of  the 
three  conductors,  the  cross-section  of  the  return  wire  must  be 
V2  times  that  of  each  of  the  other  two  wires. 

The  two  phases  hi  Fig.  30  are  sometimes  electrically  inter- 
connected at  their  middle  points,  as  shown  in  Fig.  33  at  the  left. 


FIG.  33.    A  star-connected  quarter-phase  system  to  the  left,  a  mesh-connected 
system  to  the  right. 

This  is  done  in  order  to  fix  the  difference  of  potential  between  the 
two  phases.  If  the  voltage  between  A  and  B  is  E,  then  the  volt- 
age between  the  common  point  0  and  each  wire  is  \  E,  and  the 
voltage  between  the  two  phases  is  equal  to  \  E  V2  =  E/V2.  For 
example,  the  voltage  Eda  between  D  and  A  equals  Ea  —  Ed.  These 
relations  are  shown  vectorially  in  Fig.  34.  This  circuit  is  some- 
times called  the  star-connected  quarter-phase  system. 

The  four  windings  of  a  generator  or  motor  are  sometimes  con- 
nected in  mesh,  as  indicated  in  Fig.  33  to  the  right.     With  the 


102 


THE  ELECTRIC  CIRCUIT 


[ART.  35 


star  connection,  the  star  voltages  OA,  OB,  etc.,  are  induced 
directly,  while  the  mesh  voltages  AD,  DB,  etc.,  are  established  by 
the  combination  of  the  star  voltages.  With  the  mesh  connection 
of  the  windings,  however,  the  mesh  or  line  voltages  are  induced 
directly.  The  mesh  and  star  voltages  are  shown  in  Fig.  34. 
Electrically  the  two  arrangements  are  equivalent,  provided  that 
the  proper  numbers  of  turns  are  used  in  the  windings. 

The  line  and  mesh  currents  are  indicated  in  Fig.  34.     The  line 
currents  and  those  in  the  star-connected  windings  are  represented 


FIG.  34.    A  vector  diagram  of  currents  and  voltages  in  the  quarter-phase 
system  shown  in  Fig.  33. 

by  the  sides  of  the  square,  each  current  lagging  by  the  angle  0 
with  respect  to  the  corresponding  star  voltage;  the  angle  of  lag 
depends  upon  the  character  of  the  load.  In  Figs.  33  and  34  the 
voltages  are  taken  in  the  cyclic  order  AC,  CB,  BD,  DA;  hence,  it 
is  natural  to  take  the  positive  direction  of  the  current  in  the  same 
way.  With  the  arrows  in  Fig.  33  showing  the  positive  directions 
of  the  currents,  each  line  current  is  the  difference  between  two 
adjacent  mesh  currents.  Hence,  in  the  vector  diagram  the  mesh 


CHAP.  X] 


POLYPHASE  SYSTEMS 


103 


currents  are  represented  by  the  radii  from  the  center  to  the  vertices 
of  the  -current  square.  It  will  be  seen  that  the  angle  between  the 
mesh  currents  and  the  mesh  voltages  is  also  equal  to  <f>.  While 
the  mesh  voltages  are  V2  times  as  large  as  the  star  voltages,  the 
mesh  currents  are  1/V2  times  the  star  currents.  This  condition 
is  necessary  in  order  to  have  the  same  power  per  phase  in  the 
mesh  and  star-connected  systems. 

36.  Three-phase   Y-connected    System.      This    system   is 
shown   in   Fig.  35;   the  current  and  voltage  relations  are  repre- 


Fio.  35.    A  three-phase  Y-  or  star-connected  system. 

sented  in  Fig.  36.  OA,  OB,  and  OC  represent  three  generator 
windings;  O'A',  O'B',  and  O'C'  are  the  windings  of  a  receiving 
apparatus  —  for  instance,  an  in- 
duction motor.  The  three  gener- 
ator windings  are  placed  on  the 
armature  core  at  angles  of  120 
electrical  degrees  with  respect  to 
each  other,  so  that  the  alternating 
voltages  induced  in  these  windings 
are  displaced  in  phase  by  one- 
third  of  a  cycle,  the  positive  di- 
rection in  the  windings  being 

outward.      They  are   represented 

„      r,          3    r,    •      Fio.  36.    The  line  and  star  voltages, 
by  the  vectors  Ea,  Eb,  and  Ec  in      and  the  ljne  currentg  in  the  y. 


connected  system  shown  in  Fig .  35 . 


Fig.  36.      The  motor  windings  are 

similarly   displaced,   so    that   the 

whole  system  is  symmetrical  with  respect  to  the  three  phases. 

The  currents  lag  behind  the  voltages  by  an  angle  <f>  depending 

upon  the  relative  amounts  of  resistance,  reactance,  and  counter- 

e.m.f.  in  the  circuit. 


104  THE  ELECTRIC  CIRCUIT  [ART.  36 

The  diagram  of  connections  shown  in  Fig.  35  is  also  called  the 
star  connection,  and  the  points  0  and  0'  are  called  the  neutral 
points  of  the  system.  The  voltages  between  the  line  conductors 
and  the  neutral  points  are  called  the  star  or  phase  voltages,  as 
distinguished  from  the  line  voltages,  or  voltages  between  any 
two  line  conductors.  A  line  voltage,  for  instance  between  the 
points  A  and  B,  is  equal  to  the  geometric  difference  between  the 
voltages  OA  and  OB,  as  has  been  shown  above  in  the  case  of  a 
two-phase  line.  Consequently,  the  line  voltages  are  represented 
in  Fig.  36  by  the  three  vectors  E^,  Ebc,  and  Eca,  which  connect 
the  ends  of  the  vectors  of  the  phase  voltages.  It  will  be  seen  that 
the  line  voltages  are  V3  times  as  large  as  the  phase  voltages. 

When  the  three  phases  are  perfectly  balanced  and  the  currents 
are  nearly  sinusoidal,  the  two  neutral  points  0  and  0'  may  be 
connected  by  a  wire,  as  shown  by  the  dotted  line,  or  grounded, 
and  very  little  current  will  flow  through  this  connection.  The 
reason  is  that  the  algebraic  sum  of  the  three  currents  flowing 
towards  or  from  the  neutral  points  is  equal  to  zero  at  all  instants, 
because 

sin  u  +  sin  (u  -f-  f  TT)  +  sin  (u  —  f  TT)  =0.  (175) 

This  identity  is  easily  proved  by  expanding  the  left-hand  member, 
using  the  expression  for  the  sine  of  the  sum  of  two  angles.  It 
will  also  be  seen  from  Fig.  36  that  the  geometric  sum  of  the  three 
current  vectors  is  equal  to  zero,  because  these  vectors  when  added 
form  a  closed  triangle.  In  practice,  there  are  transmission  lines 
on  which  one  or  both  neutrals  are  grounded,  although  in  some 
installations  both  neutrals  are  insulated  from  the  ground.  To 
prevent  large  currents  with  unbalanced  loads  or  during  short- 
circuits,  the  neutrals  are  often  grounded  through  protective 
resistances.  The  question  of  grounded  vs.  ungrounded  neutrals  is 
still  in  a  somewhat  controversial  stage. 

The  power  developed  in  the  generator  windings  and  available 
at  the  generator  terminals  is  3  lyEy  cos  0,  where  Ey  is  the  phase 
voltage.  Since  the  line  voltage  E±  =  EY  V3,  we  have 

P  =  3  /Y#Y  COS  0  =  7Y#A  V3  COS  <f>.          .      .       (176) 

In  practical  calculations  of  three-phase  transmission  lines  and 
electrical  machinery,  only  one  phase  is  considered;  that  is,  the 
three-phase  circuit  is  reduced  to  an  equivalent  single-phase  cir- 
cuit. Let  it  be  required,  for  example,  to  calculate  the  cross-sec- 


CHAP.  X]  ,  POLYPHASE  SYSTEMS  105 

tion  and  per  cent  voltage  regulation  of  a  three-phase  66,000-volt 
line,  to  transmit  50,000  kw.  at  80  per  cent  power-factor,  and  at 
a  loss  of  10  per  cent  of  the  useful  power;  the  spacing  to  be  1.8  m. 
First  of  all  we  find  that  the  voltage  between  each  wire  and  the 
neutral  is  66,000/V3  =  38,100  volts,  and  that  the  power  per 
phase  is  50,000/3  =  16,700  kw.  Hence,  the  problem  is  reduced 
to  the  following  one:  Determine  the  cross-section  and  per  cent 
voltage  regulation  of  a  single-phase  38,100-volt  line,  having  a 
spacing  of  1.8  m.,  the  i2r  loss  in  one  conductor  being  1670  kw., 
and  the  resistance  of  the  return  conductor  being  negligible.  The 
solution  of  this  problem  is  given  in  Art.  33  above.  A  drop  of  say 
5  per  cent  in  the  phase  or  star  voltage  means  also  a  drop  of 
5  per  cent  in  the  line  voltage,  because  of  the  fixed  ratio  1/V3 
between  the  two. 

Prob.  1.  Assuming  the  reference  axis  in  Fig.  36  to  be  horizontal,  the 
line  voltage  equal  to  44  kv.,  the  current  per  phase  73  amp.,  and  the  angle  <f> 
equal  to  15  degrees,  write  down  the  complex  expressions  for  all  the  cur- 
rents and  voltages. 

Ans.  Eb  =  22  -  j  12.7  kv. ;  E*  =  22  -  j  38. 1  kv. ;  /«  =  18.9  +  j  70.5 
amp. 

Prob.  2.  A  three-phase  60-cycle  line  is  16  km.  long;  the  spacing  be- 
tween the  wires  is  symmetrical  and  is  equal  to  61  cm.,  the  conductors  con- 
sisting of  copper  wire  of  14  mm.  diameter.  It  is  required  to  maintain 
a  voltage  of  6700  between  the  conductors  at  the  receiver  end  of  the  line. 
What  is  the  generator  voltage  when  the  load  is  equal  to  1000  kw.  at  unity 
power-factor?  Ans.  7040. 

Prob.  3.  Show  that  a  three-phase  transmission  line  may  be  treated 
as  a  single-phase  line  which  transmits  one-half  the  power  at  the  same 
voltage.  The  three-phase  line  requires  three  conductors  of  the  same  size 
as  the  single-phase  line,  with  the  same  spacing  (25  per  cent  saving  in 
material) . 

Prob.  4.  When  the  phase  currents  have  higher  harmonics,  show  that 
equalization  currents  must  flow  through  the  neutral  connection,  even 
though  the  phases  are  perfectly  balanced.  What  happens  when  the 
neutrals  are  insulated  from  each  other? 

Prob.  5.  Show  that  the  line  voltage  cannot  have  the  third,  the  ninth, 
the  fifteenth,  etc.,  harmonics,  even  if  these  harmonics  are  present  in  the 
phase  voltages. 

37.  Three-phase  Delta-connected  System.  This  method  of 
three-phase  connection  is  shown  in  Fig.  37,  one  end  of  the  line 
being  connected,  for  instance,  to  an  alternator,  the  other  end  to  a 
motor  or  to  three  transformers.  Fig.  38  represents  the  current 


100 


THE  ELECTRIC  CIRCUIT 


[AHT.  37 


and  voltage  relations.     The  currents  in  the  windings  are  differ- 
ent from  those  in  the  line.     With  the  positive  directions  of  the 


FIG.  37.    A  three-phase  delta-  or  mesh-connected  system. 

currents  indicated  in  Fig.  37,  each  line  current  is  equal  to  the 
difference  between  the  two  adjacent  currents  in  the   "  delta," 


Fia.  38.    The  voltages  and  currents  in  the  delta-connected  system 
shown  in  Fig.  37. 

Hence,  in  the  vector  diagram  the  line  or  "  star  "  currents  are 
represented  by  a  triangle,  and  the  delta  currents  by  the  rays  from 
the  center  to  the  vertices  of  the  triangle.  It  will  be  seen  that  the 
delta  currents  are  equal  to  1/V3  of  the  line  .currents,  and  are 
displaced  in  phase  by  30  degrees  with  respect  to  them.  This  also 
follows  from  the  identity 

/  sin  u  -  I  sin  (u  +  120°)  =  \/3  /  sin  (u  -  30°).  .     (177) 


CHAP.  X]  POLYPHASE  SYSTEMS  107 

While  there  are  no  neutral  points  with  a  delta  system,  one  or  more 
of  them  may  be  artificially  created  by  connecting  three  resistances 
or  reactances  in  star  as  shown  in  Fig.  37  by  dotted  lines.  The 
.Y- voltages  between  this  neutral  and  the  line  conductors  are 
shown  in  Fig.  38  by  the  vectors  Ea,  -E&  and  Ec.  The  delta  volt- 
ages are  \/3  times  as  large  as  the  star  voltages,  while  the  star  or 
line  currents  bear  the  same  ratio  to  the  delta  currents.  This  is 
necessary  in  view  of  the  power  relation 

P  =  3  #A/A  COS  </>  =  3  #Y/Y  COS  <£.        .       .       .       (178) 

In  design  and  performance  calculations  one  phase  only  is 
considered,  the  three  phases  being  identical  when  the  load  is 
balanced.  As  far  as  the  line  is  concerned,  the  delta-connected 
generator  and  load  may  be  replaced  by  equivalent  star-connected 
windings  to  give  the  same  line  currents  and  voltages.  Then  the 
line  is  designed  and  its  performance  calculated  the  same  as  in  the 
preceding  article.  As  a  matter  of  fact,  for  line  calculations  it  is 
only  necessary  to  know  the  power,  the  voltage,  and  the  power- 
factor  of  the  load.  The  fact  that  the  generator  or  the  load  is 
delta-  or  Y-connected  has  no  bearing  upon  the  line  performance 
with  a  balanced  load. 

Prob.  1.  A  2000-kw.  6600-volt  induction  motor  is  fed  from  a  66,000- 
volt  three-phase  line  through  three  step-down  transformers,  the  high- 
tension  windings  of  which  are  connected  in  Y,  the  low-tension  windings 
in  delta.  What  are  the  currents  in  these  windings  when  the  motor  is 
carrying  a  25  per  cent  overload?  It  is  estimated  that  at  this  overload 
the  power-factor  is  90  per  cent  and  the  efficiency  92  per  cent.  The  mag- 
netizing current  of  the  transformers  is  negligible. 

Ans.    26.4  and  153  amp. 

Prob.  2.  Show  that,  while  the  instantaneous  electrical  output  of  a 
single-phase  alternator  varies  at  double  the  frequency  of  the  current, 
the  output  of  a  polyphase  machine  is  practically  constant  as  long  as  the 
load  remains  constant.  Show  that  the  same  is  true  for  motors. 

Note:  For  the  electrical  relations  in  two-  and  three-phase  systems 
with  unbalanced  loads,  and  also  for  the  theory  of  the  V  and  T  connections, 
see  the  author's  Experimental  Electrical  Engineering,  Vol.  2,  Chapter  25. 
A  more  exhaustive  treatment  will  also  be  found  in  his  investigation 
entitled  Ueber  mehrphasige  Stromsysteme  bei  ungleichmdssiger  Belastung 
(published  by  Enke,  1900).  See  also  the  chapters  on  polyphase  systems 
in  Dr.  Steinmetz's  Alternating-current  Phenomena. 


CHAPTER  XI 
VOLTAGE  REGULATION  OF  THE  TRANSFORMER 

38.  Imperfections  in  a  Transformer  Replaced  by  Equivalent 
Resistances  and  Reactances.  The  reader  is  familiar  in  general 
with  the  construction  and  operation  of  the  constant-potential 
transformer  (Fig.  39).  It  consists  of  an  iron  core  upon  which 
two  windings  are  placed  as  closely  as  possible  to  each  other. 
When  one  winding  is  connected  to  a  constant-potential  alternat- 
ing-current source  of  power,  an  alternating  magnetic  flux  is  excited 
in  the  iron  core  and  an  alternating  voltage  is  induced  in  the  other 
winding.  If  this  latter  winding  is  connected  to  an  electrical  load, 
an  alternating  current  flows  through  it,  and  causes  a  correspond- 
ing flow  of  current  through  the  first  winding,  in  order  that  power 
may  be  transmitted  from  the  primary  into  the  secondary  circuit. 

The  constant-potential  transformer  is  one  of  the  most  perfect 
pieces  of  electrical  apparatus,  in  that  its  efficiency  (in  medium  and 
large  sizes)  is  nearly  one  hundred  per  cent,  and  its  voltage  regula- 
tion with  varying  load  is  quite  close.  On  the  other  hand,  the 
requirements  for  voltage  regulation  are  quite  exacting,  there  being 
no  provision  in  the  apparatus  itself  for  adjusting  the  voltage,  like 
the  field  rheostat  in  a  generator.  Therefore,  the  pre-determina- 
tion  of  the  voltage  regulation  of  a  transformer  is  of  considerable 
.practical  importance. 

Numerically,  the 'regulation  of  a  transformer  is  expressed  in 
a  manner  similar  to  that  given  in  Art.  33  above  for  the  trans- 
mission line.  Let,  for  example,  the  rated  secondary  voltage  of  a 
ten-to-one  transformer  be  220  volts,  and  let  us  suppose  that  a 
primary  e.m.f.  of  2280  volts  is  necessary  in  order  to  have  the 
rated  secondary  voltage  at  the  rated  load.  Let  now  the  second- 
ary circuit  be  opened ;  the  secondary  voltage  will  rise  to  prac- 
tically 228  volts,  provided  that  the  primary  voltage  is  kept  con- 
stant. Then,  by  definition,  the  regulation  of  the  transformer  at 
this  load  is  100  (228  -  220) /220  =  3.64  per  cent.  Let,  in  general, 
108 


CHAP.  XI]        REGULATION  OF  THE  TRANSFORMER  109 

the  secondary  terminal  voltage  at  a  certain  load  be  E2,  that  at  no 
load  EM  Then,  by  definition, 

per  cent  regulation  =  100  (#02  -  Ed/E*    .     .     (179) 

The  difference  between  the  no-load  voltage  and  that  at  full  load 
is  due  to  slight  imperfections  in  the  transformer  itself.  There- 
fore, in  order  to  be  able  to  calculate  the  voltage  regulation  at  a 
given  load,  it  is  necessary  to  learn  the  nature  of  these  imperfec- 
tions; for  purposes  of  computation,  it  is  convenient  to  replace 
these  imperfections  by  certain  resistances  and  reactances,  as  shown 
in  Fig.  39. 

In  an  ideal  transformer  the  ratio  of  the  primary  to  the  second- 
ary voltage  is  equal  to  the  ratio  of  the  numbers  of  turns  in  the 
corresponding  windings.  The  same  relation  is  very  nearly  true  in 
any  good  transformer  at  no  load.  This  follows  from  the  fact  that 
the  two  windings  are  linked  with  the  same  magnetic  flux,  and 
hence  the  voltage  induced  per  turn  is  the  same  in  both.  Hence, 
denoting  the  primary  and  secondary  induced  voltages  by  En  and 
Eiz,  and  the  corresponding  numbers  of  turns  in  series  by  n\  and  HZ, 
we  have 

Eil/Ei^  =  nl/nz  .......     (180) 

Furthermore,  in  an  ideal  transformer 

7ini  =  /2n2,  .     ......     (181) 

that  is,  the  currents  are  inversely  as  the  numbers  of  turns,  or  the 
primary  ampere-turns  are  equal  and  opposite  to  the  secondary 
ampere-turns.  This  is  because  an  ideal  transformer  is  supposed 
to  have  no  reluctance  in  its  magnetic  circuit,  so  that  no  ampere- 
turns  are  required  to  maintain  a  magnetic  flux  in  it.  Consequently, 
any  secondary  current  required  by  the  load  automatically  draws 
a  compensating  primary  current  from  the  source  of  power,  of  such 
value  that  eq.  (181)  is  satisfied.  In  a  real  transformer  the  primary 
ampere-turns  are  slightly  different  from  the  secondary  ampere- 
turns,  and  the  difference  between  the  two  is  just  sufficient  to  main- 
tain the  alternating  flux  through  the  reluctance  of  the  core,  and  to 
supply  the  core  loss.  Multiplying  eqs.  (180)  and  (181)  term  by 
term,  and  canceling  n\  and  n2,  we  find  that 


This  simply  means  that  an  ideal  transformer  transmits  power 
from  the  primary  into  the  secondary  circuit  without  loss. 


110 


THE   ELECTRIC   CIRCUIT 


[ART.  38 


(a)  The  Ohmic  Drop.  One  of  the  causes  of  the  internal  volt- 
age drop  in  a  transformer  is  the  ohmic  resistance  of  its  windings. 
Because  of  the  resistance  of  the  primary  winding,  the  primary 
terminal  voltage  El  (Fig.  39)  is  slightly  larger  than  the  induced 

I, 


Iron  Cure  Windings 

FIG.  39.     Imperfect  ions  in  a  transformer  represented   by   resistances  and 
reactances. 

counter-e.m.f.  EH  which  balances  it.  The  secondary  resistance 
causes  a  voltage  drop,  so  that  the  secondary  terminal  voltage  E2 
is  smaller  than  the  secondary  induced  e.m.f.  Ei2.  Thus,  the 
effect  of  the  internal  resistances  upon  the  terminal  voltages  is 
such  as  to  make  the  ratio  E*/Ei  smaller  than  the  ratio  n2/ni. 
The  windings  themselves  may  be  thought  of  as  devoid  of  resist- 
ance, but  corresponding  resistances  r\  and  r2  may  be  placed  out- 
side the  transformer,  as  shown  in  Fig.  39.1 

(b)  The  Reactive  Drop.  Another  imperfection  or  cause  of 
internal  voltage  drop  is  the  so-called  leakage  reactance  of  the 
windings.  The  total  magnetic  flux  in  a  loaded  transformer  may 
be  considered  as  consisting  of  three  components;  viz.,  the  useful 
flux  linking  with  both  the  primary  and  the  secondary  windings, 
the  primary  leakage  flux  linking  with  the  primary  winding  only, 
and  the  secondary  leakage  flux  linked  with  the  secondary  winding 
only.  In  an  ideal  transformer  the  two  last-named  fluxes  are 
absent  because  the  two  windings  are  supposed  to  be  perfectly 
interwoven,  so  as  to  leave  no  room  for  the  leakage  flux.  The 
primary  leakage  flux,  being  produced  by  the  primary  current,  is 
in  phase  with  it,  and  induces  an  e.m.f.  in  lagging  quadrature  with 

1  The  equivalent  resistances  TI  and  r2  must  replace  not  only  the  true 
ohmic  resistances  of  the  windings,  but  should  also  account  for  the  eddy- 
currcnt  loss  in  the  conductors.  In  low-tension  windings  made  of  heavy 
conductors,  this  latter  loss  may  be  at  least  as  great  as  the  theoretical  iV  loss. 
In  new  transformers  the  eddy-current  loss  can  only  be  estimated;  in  actu- 
ally built  transformers  it  is  calculated  from  the  wattmeter  reading  on  short 
circuit. 


CHAP.  XI]        REGULATION  OF  THE  TRANSFORMER  111 

this  current.  This  e.m.f.  must  be  balanced  by  part  of  the  applied 
primary  voltage,  so  that  either  this  voltage  or  the  induced  e.m.f. 
EH  must  be  different  from  that  in  an  ideal  transformer.  The 
effect  of  the  secondary  leakage  reactance  is  similar,  in  that  it 
absorbs  part  of  the  secondary  induced  voltage  Ei2,  and  makes 
the  secondary  terminal  voltage  Ez  different  from  that  in  an 
ideal  transformer.  It  is  shown  below  that,  with  a  load  of  lag- 
ging power-factor,  the  reactive  drop  in  both  windings  lowers  the 
secondary  terminal  voltage.  With  a  leading  secondary  current, 
the  reactive  drop  is  hi  such  a  phase  position  as  to  raise  the  sec- 
ondary voltage.  For  purposes  of  computation,  the  transformer 
windings  are  assumed  to  produce  no  magnetic  leakage  fluxes,  but 
imaginary  reactance  coils  are  connected  in  series  with  the  wind- 
ings (Fig.  39).  The  reactances  x\  and  Xz  of  these  coils  are  such  as 
to  cause  the  same  reactive  voltage  drop  as  that  due  to  the  actual 
leakage  fluxes  in  the  transformer.1 

(c)  The  Exciting  Admittance.  Having  thus  made  the  windings 
of  the  transformer  perfect  by  placing  their  impedances  outside, 
we  still  have  the  problem  of  making  the  magnetic  circuit  ideal 
also.  As  stated  before,  the  primary  and  secondary  ampere-turns 
are  not  quite  equal,  because  of  a  certain  number  of  ampere-turns 
necessary  to  magnetize  the  iron.  This  means  that  a  current 
must  flow  through  the  primary  winding  even  when  the  secondary 
circuit  is  open.  This  current  is  called  the  no-load  or  magnetizing 
current  of  the  transformer.  Its  amount  depends  upon  the  reluct- 
ance of  the  magnetic  circuit  and  upon  the  core  loss  (hysteresis 
and  eddy  currents).  For  purposes  of  computation  the  iron  core 
may  be  assumed  to  be  of  zero  reluctance,  and  to  have  no  core 
loss;  but  we  may  imagine  a  fictitious  or  equivalent  susceptance 
60  and  a  conductance  g0  (Fig.  39)  connected  across  the  primary 
winding  to  draw  a  current  equal  hi  phase  and  magnitude  to  the 
exciting  current  of  the  transformer.  Both  go  and  60  are  shown 
connected  across  the  induced  voltage  EH,  because  both  the  mag- 
netizing current  and  the  core  loss  depend  upon  the  value  of  the 
flux  and  consequently  upon  the  value  of  En,  which  is  proportional 
to  the  flux.  Let  the  calculated  or  measured  core  loss  be  equal  to 
Po  watts;  then  g0  is  determined  from  the  equation 

P0  =  En*go (182) 

1  For  further  details  in  regard  to  the  leakage  reactance  of  transformers, 
see  the  author's  Magnetic  Circuit,  Art.  64. 


112  THE  ELECTRIC   CIRCUIT  [ART.  39 

The  pure  magnetizing  current,  without  the  core-loss  component, 
is  in  phase  with  the  flux  which  it  produces,  and  therefore  is  in 
quadrature  with  the  induced  voltage  En.  For  this  reason,  it  is 
represented  as  flowing  through  a  pure  susceptance.  Knowing 
the  pure  magnetizing  current  IQ  ',  the  susceptance  b0  is  determined 
from  the  equation l 

bo  =  I07En (183) 

Neither  the  core-loss  component  nor  the  pure  magnetizing  current 
are  proportional  to  the  flux  or  to  the  voltage  EH,  so  that  strictly 
speaking  both  b0  and  g0  are  functions  of  the  counter-e.m.f.  EH. 
However,  in  practice,  EH  varies  so  little  with  the  load  that  it  is 
admissible  to  assume  g0  and  &o  to  be  constant  quantities.  More- 
over, the  influence  of  the  magnetizing  current  upon  the  voltage 
regulation  is  negligible  in  most  cases.  The  magnetizing  current 
is  mentioned  here  only  for  the  sake  of  completeness,  so  as  to  make 
the  transformer  core  absolutely  perfect.  We  shall  see  in  the  next 
chapter  that  60  and  g0  are  of  considerable  importance  in  the  per- 
formance of  the  induction  motor. 

Thus,  by  the  foregoing  reasoning,  both  the  core  and  the  wind- 
ings of  the  transformer  are  made  ideal,  and  all  the  imperfections 
are  replaced  by  external  resistances  and  reactances.  Having  done 
this,  the  performance  of  a  transformer  can  be  readily  treated  either 
graphically  or  analytically,  as  explained  below. 

Prob.  1.  Draw  a  diagram  similar  to  Fig.  39  for  a  transformer  with 
several  secondary  windings  supplying  independent  load  circuits. 

Prob.  2.   Draw  a  diagram  similar  to  Fig.  39  for  an  auto-transformer. 

39.  The  Vector  Diagram  of  a  Transformer.  Having  re- 
duced the  transformer  to  an  equivalent  electric  circuit  (with  a 
perfect  magnetic  link),  the  current  and  voltage  relations  at  a 
certain  load  may  be  represented  by  a  vector  diagram  (Fig.  40). 
In  order  to  make  the  relations  clearer,  the  voltage  drop  and  the 
losses  are  greatly  exaggerated.  For  this  reason,  the  graphical 
treatment  is  more  suitable  for  purposes  of  explanation  than  for 
numerical  computations.  For  actual  calculations  the  analytical 
method  given  in  the  next  article  is  preferable. 

The  calculation  of  the  core  loss  and  of  the  magnetizing  current  belongs 
properly  to  the  theory  of  magnetic  phenomena,  and  is  treated  in  detail  in  the 
author's  Magnetic  Circuit,  Arts.  19,  33,  and  34.  Here  the  values  of  P0  and 
/o'  are  supposed  to  be  known. 


CHAP.  XI]        REGULATION  OF  THE  TRANSFORMER 


113 


Let  the  secondary  terminal  voltage  and  the  load  be  given,  so 
that  the  vectors  E2  and  72  can  be  drawn  in  magnitude  and  rela- 
tive phase  position.  The  secondary  induced  e.m.f.  Ei2  is  found 


FIG.  40.    The  vector  diagram  of  a  transformer. 

by  adding  to  E2  the  ohmic  drop  72r2  in  phase  with  72,  and  the 
reactive  drop  I2Xz  in  leading  quadrature  with  72. 

The  primary  induced  voltage  En  is  in  phase  with  Eiz,  because 
both  are  induced  by  the  same  magnetic  flux.  The  magnitudes 
of  the  two  voltages  are  as  the  respective  numbers  of  turns;  see 


114  THE  ELECTRIC  CIRCUIT  [ART.  39 

eq.  (180).  The  vector  marked  En  in  Fig.  40  is  in  reality  equal  and 
opjx>site  to  En,  and  represents  the  part  of  the  primary  terminal 
voltage  that  balances  EH.  Without  the  primary  drop,  the  total 
applied  primary  voltage  would  be  equal  to  En.  But,  on  account 
of  the  primary  drop  in  the  transformer,  the  applied  voltage  Ei  is 
obtained  by  adding  to  En  the  resistance  drop  /irx  in  phase  with 
the  primary  current  /i  and  the  reactive  drop  IiXi  in  leading 
quadrature  with  I\. 

In  order  to  be  able  to  construct  the  vectors  /in  and  IiXi  it 
is  necessary  to  know  the  vector  of  the  primary  current  /i  in 
magnitude  and  phase  position.  In  an  ideal  transformer,  the 
primary  current  is  in  exact  phase  opposition  to  the  secondary 
current,  and  the  ratio  of  the  two  currents  is  inversely  as  the  ratio 
of  the  respective  numbers  of  turns;  see  eq.  (181).  In  the  actual 
transformer,  the  primary  current,  in  addition  to  this  component 
7  2  (nz/ni)  "  transmitted  into  the  secondary,"  has  a  magnetizing 
component  70,  which  serves  to  maintain  the  alternating  flux  in 
the  core,  and  which  is  not  transmitted  into  the  secondary  circuit. 
The  total  primary  current  is  the  geometric  sum  of  the  two  com- 
ponents, and  can  be  constructed  if  the  magnetizing  current  70  is 
known. 

The  magnetizing  current  itself  consists  of  two  components,  as 
explained  in  the  preceding  article,  under  (c).  One  component, 
/</,  is  in  phase  with  the  useful  magnetic  flux  0,  and  would  be  the 
only  magnetizing  component  if  the  iron  had  no  hysteresis  and  no 
eddy  currents.  This  component  is  in  quadrature  with  the  induced 
voltage  En,  and  is,  with  respect  to  it,  the  reactive  component  of 
the  magnetizing  current.  The  other  component,  /</',  in  phase 
with  En,  represents  a  loss  of  power,  and  is  therefore  called  the 
energy  or  loss  component  of  the  magnetizing  current.  Knowing 
/o'  and  /o",  the  vector  70  is  easily  obtained. 

The  vector  of  flux,  0,  is  drawn  in  leading  quadrature  with  the 
induced  e.m.f.  #,-2,  in  accordance  with  Faraday's  law  of  induction. 
If  the  flux  varies  according  to  the  law 


(184) 
the  induced  e.m.f.  varies  according  to  the  law 

e<2  =  -  n2  dd>t/dt  =-2  ^mrh.  cos  2  irft,      .     .     (185) 
the  second  sine-wave  lagging  by  90  degrees  behind  the  first. 


CHAP.  XI]         REGULATION  OF  THE  TRANSFORMER  115 

It  is  assumed  in  the  construction  of  Fig.  40  that  the  primary  and 
secondary  inductive  drops  can  be  calculated  separately.  Such  is, 
however,  not  the  case  with  our  present  state  of  knowledge;  both 
theory  and  experiment  enable  us  to  determine  only  the  total  re- 
active drop,  including  primary  and  secondary.  Therefore,  when 
it  is  desired  to  use  the  vector  diagram  for  actual  computations,  it 
is  customary  to  ascribe  one  half  of  the  Ix  drop  to  the  primary  and 
the  other  half  to  the  secondary  circuit. 

Usually,  the  magnetizing  component  of  the  primary  current 
can  be  neglected;  then  it  does  not  make  any  difference  how  the 
inductive  drop  is  distributed.  It  will  be  shown  in  the  next  article 
that  in  such  case  the  voltage  regulation  depends  only  upon  the 
total  impedance  drop,  either  calculated  or  determined  from  a 
short-circuit  test.  When  the  internal  voltage  drop  is  given  in 
per  cent,  it  is  understood  to  refer  to  the  no-load  voltage  of  each 
particular  circuit.  For  instance,  if  the  reactive  voltage  drop  in 
a  20/1-kv.  transformer  is  said  to  be  5  per  cent,  this  means  that 
the  secondary  drop  is  2.5  per  cent  of  1000  volts,  or  is  equal  to  25 
volts,  and  that  the  primary  drop  is  2.5  per  cent  of  20,000  volts,  or 
is  equal  to  500  volts. 

Prob.  1.  What  is  the  regulation  of  a  600-kw.,  2200/220- volt,  25-cycle 
transformer  at  the  rated  current  and  at  80  per  cent  power-factor  (lag- 
ging)? The  total  reactive  drop  is  10  per  cent,  the  primary  ohmic  drop 
is  2.2  per  cent,  and  the  secondary  ohmic  drop  2.8  per  cent.  The  mag- 
netizing current  may  be  neglected.1  Ans.  10.1  per  cent. 

Prob.  2.  Determine  the  per  cent  voltage  regulation  of  the  trans- 
former specified  in  the  preceding  problem  at  the  rated  load  and  at  80  per 
cent  power-factor,  leading. 

Ans.  — 1.4  per  cent.  The  negative  sign  indicates  a  rise  in  secondary 
voltage,  instead  of  a  drop. 

Prob.  3.  Correct  the  vector  diagram  of  problem  1  for  the  magnetiz- 
ing current,  knowing  that  the  core  loss  amounts  to  20  kw.,  and  that  8500 
effective  ampere-turns  are  necessary  to  maintain  the  flux,  without  the  iron 
loss.  The  number  of  turns  in  the  secondary  winding  is  64. 

Prob.  4.   Adapt  the  diagram  shown  in  Fig.  40  to  an  auto-transformer. 

40.  Analytical  Determination  of  Voltage  Regulation. — Ap- 
proximate Solution.  As  explained  above,  it  is  preferable  to 
calculate  the  voltage  regulation  of  a  transformer  analytically, 

1  An  excessive  internal  drop  is  selected  purposely  to  enable  the  student 
to  construct  an  accurate  vector  diagram  to  a  convenient  scale.  The  losses 
and  the  magnetizing  current  in  problem  3  also  are  too  high  for  a  standard 
transformer. 


116  THE  ELECTRIC  CIRCUIT  [ART.  40 

because  the  vectors  of  voltage  drop  are  very  small  as  compared 
with  those  of  the  primary  and  secondary  voltages.  The  relations 
shown  in  Figs.  39  and  40  are  expressed  analytically  by  the 

equations 

Et=En-!&, (186) 

E^Eu+ItZi (187) 

Since  our  purpose  is  to  find  the  relation  between  Et  and  E%,  it 
is  necessary  to  eliminate  from  these  equations  EH  and  Ei2.  The 
relation  between  EH  and  Eiz  is  given  by  eq.  (180);  therefore,  we 
multiply  eq.  (186)  by  wi/na  and  subtract  it  from  eq.  (187).  The 
result  is 

El  -  (ni/n2)#2  =  IiZj.  +  (wi/n2)/2Z2.     .     .     (188) 

The  correct  relation  between  /i  and  72  is  (Fig.  39) 

1 1  =  1 2  (n2/ni)  +  I0=  IL+IO,     .     .     .     (189) 
where 

/L  =  /t(ni/nO (190) 

is  the  primary  load  current,  or  that  part  of  the  primary  current 
which  is  transmitted  into  the  secondary  circuit.  In  a  great 
majority  of  practical  cases  the  magnetizing  current  is  only  a  few 
per  cent  of  the  total  primary  current  at  the  rated  load.  The 
voltage  drop  in  the  primary  winding  is  also  but  a  few  per  cent  of 
the  line  voltage  EI.  For  these  reasons,  it  is  permissible  in  Fig.  39 
to  transfer  the  exciting  admittance  Y0  from  the  place  MN  to  the 
primary  terminals  AB.  The  voltage  drop  in  the  transformer  is 
then  caused  only  by  the  load  current,  so  that  for  the  purpose  of 
calculating  regulation  we  may  use  the  approximate  relation 

/,  =  /L=/»(n,/n1) (191) 

Substituting  for  /i  and  72  their  values  from  eq.  (191)  in  terms  of 
I  L,  we  finally  obtain 

^i-^L=/L[Zi  +  (ni/n0^d (192) 

In  this  equation,  the  quantity 

EL  =  (ni/n2)#2 (193) 

is  called  the  primary  load  voltage,  or  the  secondary  terminal  volt- 
age reduced  to  the  primary  circuit.  The  expression  (ni/n2)2Z2  is 
called  the  secondary  impedance  reduced  to,  or  transferred  into, 
the  primary  circuit.  The  quantity 

Z  =  Zl  +  (n!/n2)2Z8 (194) 


CHAP.  XI]        REGULATION  OF  THE  TRANSFORMER  117 

is  called  the  total  or  equivalent  impedance  of  the  transformer 
reduced  to  the  primary  circuit. 

Using  in  eq.  (192)  the  abbreviated  notation  introduced  in 
eq.  (194),  we  get 

E1-EL=ILZ (195) 

Equation  (195)  corresponds  to  the  simplified  equivalent  diagram 
of  the  transformer  shown  in  Fig.  41.  This  diagram  differs  from 
Fig.  39  in  two  respects:  (1)  The  magnetic  link  is  omitted,  the 
primary  circuit  being  connected  directly  to  the  modified  second- 
ary circuit;  (2)  the  exciting  admittance  is  connected  across  the 
primary  terminal  voltage  instead  of  across  the  induced  voltage. 
The  latter  change  makes  the  equivalent  diagram  only  approxi- 
mately correct,  but  simplifies  computations  greatly. 

Equation  (195)  is  identical  in  form  with  eq.  (159),  Art.  33, 
for  the  voltage  drop  in  a  transmission  line;  both  are  solved,"  and 
the  per  cent  voltage  drop  determined,  in  the  same  way.  In  fact, 
without  the  exciting  admittance  Y0,  the  equivalent  diagram  shown 
in  Fig.  41  reduces  the  performance  of  a  transformer  to  that  of  a 
transmission  line. 

Expression  (194)  for  the  equivalent  impedance  shows  that 
resistances  and  reactances  can  be  transferred  from  the  secondary 


B  N  D 

FIG.  41.    The  approximately  equivalent  diagram  of  a  transformer  or  an 
induction  motor. 

circuit  into  the  primary,  and  vice  versa,  by  multiplying  them  by 
the  square  of  the  ratio  of  the  numbers  of  turns.  For  instance,  in 
a  10,000/1000-volt  transformer,  a  1-ohm  resistance  in  the  low- 
tension  circuit  causes  the  same  per  cent  voltage  drop  as  a  100- 
ohm  resistance  in  the  high-tension  circuit.  This  is  easily  verified 
as  follows:  Let  the  current  in  the  low-tension  circuit  be  20  amp.; 
then  in  the  high-tension  circuit  the  current  will  be  2  amp.  The 
drop  in  the  1-ohm  resistance  is  20  volts,  or  2  per  cent  of  the 
secondary  voltage.  The  drop  in  the  100-ohm  resistance  is  200 
volts,  which  is  2  per  cent  of  the  primary  voltage.  In  other  words, 
the  same  reduction  in  the  load  voltage  will  be  produced  by  using 


118  THE   ELECTRIC  CIRCUIT  [ART.  40 

cither  a  resistance  of  one  ohm  in  the  secondary  circuit  or  100  ohms 
in  the  primary  circuit. 

The  secondary  resistance  r2  transferred  into  the  primary  cir- 
cuit is  denoted  in  Fig.  41  by  r2',  where 

r,'  =  r,  (m/nO1  .......     (196) 

C  or  respo  ndingly 

a^aiCni/n,)2  .......  (196a) 

The  equivalent  impedance  Z  consists  of  the  quadrature  sum 
of  the  equivalent  resistance 

....     (197) 


and  the  equivalent  reactance 

x  =  x,  +  xz'  =  X!  +  (ni/n2)2:r2  ....  (197a) 
The  equivalent  resistance  is  easily  calculated,  knowing  the  re- 
sistances of  the  two  windings  and  the  voltage  ratio  of  the  trans- 
former. Or  else  it  is  calculated  directly  from  the  i2r  loss  measured 
by  a  wattmeter  in  a  short  circuit  test.  The  equivalent  reactance 
is  calculated  from  the  terminal  voltage  in  the  short-circuit  test, 
making  a  proper  allowance  for  the  known  resistance  drop.  To 
illustrate,  when  the  secondary  circuit  is  short-circuited,  eq.  (195) 
becomes 

El  =  ILZ  ........     (198) 

Ei  and  IL  are  measured  directly,  so  that  Z  can  be  calculated. 
Knowing  the  equivalent  resistance  r  =  P/Ii?,  the  reactance  is 
calculated  from  the  expression  x  =  Vz2  —  r2.  For  a  new  trans- 
former, the  total  equivalent  leakage  inductance  is  estimated  with 
sufficient  accuracy  by  means  of  various  semi-empirical  formulae;  l 
or  else  the  total  impedance  drop  ILZ  is  taken  as  a  certain  per- 
centage of  the  rated  voltage,  from  previous  experience  with  similar 
transformers. 

Prob.  1.  Check  analytically  the  answers  to  problems  1  and  2  in  the 
preceding  article. 

Prob.  2.  The  high-tension  winding  of  a  2000-kva.,  33/1  1-kv.  trans- 
former was  short-circuited,  and  the  voltage  on  the  low-tension  side  ad- 
justed so  as  to  circulate  the  rated  current  through  the  windings.  The 
instrument  readings  were  470  volts  and  30  kw.  Calculate  the  per  cent 
ohmic  and  reactive  drops  in  the  transformer.  Ans.  1.5  and  4  per  cent. 

Prob.  3.  Deduce  a  formula  similar  to  (192),  but  referring  to  the 
secondary  circuit. 

1  See  for  instance  the  author's  Magnetic  Circuit,  Art.  64. 


CHAP.  XI]        REGULATION  OF  THE  TRANSFORMER  119 

Prob.  4.  Show  how  the  voltage  regulation  of  a  transformer  can  be 
estimated,  using  Mershon's  diagram  given  in  various  electrical  handbooks 
and  pocketbooks. 

Prob.  5.  The  primary  voltage  of  a  given  transformer  is  kept  constant 
at  a  known  value.  Determine  the  percentage  internal  drop  (Ei  —  EL)/  Ei 
for  a  given  impedance  of  the  load.1  Solution:  Let  the  load  impedance, 
reduced  to  the  primary  circuit,  be  ZL',  then  the  load  current  is 

IL  =  Ei/(ZL  +  Z), 

where  Z  is  the  equivalent  impedance  of  the  transformer  itself,  supposed 
to  be  known.  The  load  voltage  is 

EL  =  E,-  ZIL  =  EiZL/(ZL  +  Z)  =  #!/[!  +  (Z/ZL)]. 

Having  expressed  all  the  known  and  unknown  quantities  in  the  complex 
form,  in  either  Cartesian  or  polar  coordinates,  the  magnitude  and  direc- 
tion of  EL  can  be  determined,  by  using  the  general  method,  i.e.,  equating 
the  real  and  the  imaginary  parts  on  both  sides  of  the  equation. 

Prob.  6.  The  equation  for  EL  given  in  the  preceding  problem  leads 
to  involved  numerical  computations.  Moreover,  the  difference  EI  —  EL 
cannot  be  accurately  determined  in  this  way  when  EL  differs  but  little 
from  #1.  Show  how  to  simplify  the  numerical  work,  by  taking  advan- 
tage of  the  fact  that  Z  is  small  compared  with  ZL-  Solution  :  When  a 
quantity  a  is  small  compared  to  unity,  we  have  by  division  1/(1  +  a)  = 
1  —  a  +  a2  —  etc.  We  have  accordingly 

EL  =  E1[l-  (Z/ZL)]  approximately, 
or 

EL  =  Ei  -  E^/ZL  .....    .    .....     (A) 

Let  EL  be  the  vector  of  reference;  consequently  EI  =  EI  (cos  9  +  j  sin  0). 
Let  also  ZL  =  ZL  (cos  $L  +  j  sin  <£L)  and  Z  =  z  (cos  <t>  +  j  sin  <£).  Then 
according  to  eqs.  (154)  and  (156), 

E.Z/ZL  =  E.Z/ZL  [cos  (e  +  4>  -  to)  +  j  sin  (e  +  <t>  -  to)], 
or,  denoting  EiZ/Zi  by  A^i  and  <t>  —  <f>L  by  0,  we  have 
A#j  =  A#!  [cos  (e  +  0)  +  j  sin  (e  +  0)]. 
Equation  (A)  may  now  be  written  in  the  form 

Ei  (cos  e  +  j  sin  e}  =  EL  +  A#i  [cos  (0  +  0)  +  j  sin  (e  +  ft)], 
where 


is  a  known  quantity,  as  well  as  the  angle  /3  =  #  —  $L.    Separating  the 
real  and  the  imaginary  parts,  we  get 

E1coso  =  EL+  &EiCos(e  +  p),       .........     (B) 

Ei  sin  e  =  &Ei  sin  (0  +  0)  =  &Ei  sin  0  cos  0  +  A#i  cos  6  sin  0.    (C) 

1  The  conditions  in  this  problem  differ  from  those  in  the  text  above  in 
two  respects:  (1)  The  primary  voltage  is  given  instead  of  the  secondary; 
(2)  the  load  is  given  by  its  impedance  instead  of  the  current  and  power- 
factor. 


120 


THE  ELECTRIC  CIRCUIT 


[AKT.  41 


From  eq.  (C),  dividing  both  sides  by  cos  6,  we  find 

tan  0  =  AEi  sin  P/(Ei  -  A#i  cos  0),      .     .     .     .     (D) 

from  which  0  can  be  calculated.     Using  in  eq.  (B)  the  transformation 
cos  e  =  1  —  2  sin2  £  e,  the  same  as  in  Art.  33,  we  get,  after  division  by  Ei, 

(Ei  -  EL) /EL  =  ( A#,/#,)  cos  (e  +  0)  +  2  sin2  J  0.   .    .     (E) 

While  the  derivation  of  formulae  (D)  and  (E)  may  seem  somewhat  tedi- 
ous, the  results  are  in  the  form  most  convenient  for  numerical  work. 

41.  Analytical  Determination  of  Voltage  Regulation. — Exact 
Solution.1  The  approximation  made  in  the  preceding  article 
consists  in  shifting  the  exciting  admittance  Y0  so  that  it  is  con- 
nected across  the  primary  terminal  voltage  EI,  instead  of  across 
the  primary  induced  voltage  EH  (compare  Figs.  39  and  41). 
Retaining  the  exciting  admittance  in  its  correct  place,  we  obtain 
the  equivalent  diagram  shown  in  Fig.  42.  The  secondary  im- 


B  N 

FIG.  42.     The  correct  equivalent  diagram  of  a  transformer  or  an 
induction  motor. 

pedance  is  reduced  to  the  primary  circuit  as  before,  by  being 
multiplied  by  the  square  of  the  ratio  of  turns  (ni/n2)2.  This  pro- 
cedure is  strictly  correct,  the  magnetic  link  being  by  assumption 
perfect. 

Equations  (186)  and  (187)  hold  here  as  before,  but  instead 
of  using  the  approximate  eq.  (191)  we  shall  use  the  correct  rela- 
tion (189).  The  magnetizing  current  is 

Jo=#iiF0, (199) 

so  that  the  total  primary  current  is 

/!=  /L+^-iFo (200) 

Expressing  1^  and  72  in  eqs.  (186)  and  (187)  through  the  load 
current  /  L,  and  eliminating  EH  and  Ei2  as  before,  we  obtain 

$1  ~  Z!/L)/(I  +  ZiFo)  =  EL  +  IL  (m/n,)2Z,.  (201) 
This  equation  takes  the  place  of  the  approximate  eq.  (192).  The 
two  equations  become  identical  when  F0  =  0. 

1  This  article  may  be  omitted  if  desired. 


CHAP.  XI]         REGULATION  OF  THE  TRANSFORMER  121 

The  complex  quantity  1  +  ZiY0  which  enters  into  eq.  (201) 
may  be  called  a  correction  factor,  and  may  be  represented  in  the 
form 

K=  l+Z1F0  =  A;(cosa+jsina).  .     .     .     (202) 

Since  the  ratio  of  two  complex  quantities  is  also  a  complex  quan- 
tity, eq.  (201)  may  be  expressed  in  the  form 

Ecl=EL  +  ILZc, (203) 

where  the  corrected  primary  voltage 

Ed  =  Ei/K  =  (Ei/k)  [cos  (6  -  a)  +  j  sin  (8  -  a)],  (204) 
and  the  corrected  equivalent  impedance 

Zc  =  Zi/K  +  (n./n^Z,  =  zc  (cos  ^  +  j  sin  ^) .       .     (205) 

Equation  (203)  is  of  the  same  standard  form  as  eqs.  (195)  and 
(159),  and  can  be  solved  by  the  method  given  in  Art.  33. l 

This  problem  can  be  solved  also  by  keeping  the  complex  quan- 
tities in  the  orthogonal  form.  The  student  will  profit  by  working 
out  the  details  for  himself. 

Sometimes  it  is  desired  to  know  the  voltage  regulation  of  a  trans- 
former over  a  certain  range  of  loads,  the  results  being  represented 
in  the  form  of  a  curve.  In  such  a  case,  it  makes  no  difference 
for  which  particular  loads  the  regulation  is  actually  calculated, 
provided  that  these  loads  are  selected  within  certain  limits. 
If  the  primary  voltage  is  given  and  is  constant,  it  may  be  more 
convenient  to  perform  the  calculations  (according  to  Fig.  42), 
not  for  an  assumed  current  /2,  but  for  an  assumed  load  imped- 
ance ZL.  Combining  the  impedances  in  series  and  the  admit- 
tances in  parallel,  the  whole  circuit  connected  at  the  primary 
terminals  is  finally  reduced  to  one  impedance.  Dividing  the 
primary  voltage  by  this  impedance  gives  the  primary  current,  and 
consequently  the  drop  in  the  primary  impedance  Zi.  Thus,  the 
voltage  EH  becomes  known,  and  the  current  70  can  be  calculated. 
After  this,  the  current  I  L  and  the  drop  / z,Z2  are  calculated.  This 
drop,  being  subtracted  from  En,  gives  the  desired  secondary 
voltage  Ei,  reduced  to  the  primary  circuit. 

1  Namely,  when  I L  =  0,  EL  =  Eci,  so  that  per  cent  regulation  is  equal  to 
lQO&E/(Eci  —  &E),  where  AE  =  Eci  —  EL  (algebraically,  not  geometrically). 
Consequently,  eq.  (166)  and  the  table  on  page  96  are  directly  applicable. 


CHAPTER  XII 

PERFORMANCE   CHARACTERISTICS   OF   THE 
INDUCTION   MOTOR 

42.  The  Equivalent  Electrical  Diagram  of  an  Induction  Mo- 
tor. The  student  is  supposed  to  be  familiar  with  the  general  (quali- 
tative) explanation  of  the  performance  of  a  polyphase  induction 
motor,  and  with  the  general  shape  of  the  load  characteristics.1 
It  will  be  shown  here  how  to  predetermine  the  performance  char- 
acteristics of  a  given  induction  motor  by  reducing  it  to  an  equiva- 
lent electric  circuit,  similar  to  that  of  a  transformer. 

The  following  experiment  shows  the  possibility  of  such  an 
equivalent  diagram.  A  brake  test  is  performed  on  the  motor, 
and  the  primary  current  and  the  power-factor  are  plotted  against 
the  output  as  abscissae.  Then  the  rotor  is  blocked,  and  variable 
non-inductive  resistances  are  inserted  into  its  phase  windings. 
If  the  rotor  has  a  squirrel-cage  secondary,  resistances  must  be 
inserted  in  series  with  each  bar,  or  into  each  section  of  the  end- 
rings  between  consecutive  bars.  The  motor  is  thus  reduced  to  a 
polyphase  transformer,  the  inserted  secondary  resistances  repre- 
senting the  load.  A  load  test  is  performed  on  this  transformer, 
and  the  curves  of  primary  current  and  power-factor  are  plotted 
against  the  total  fir  loss  in  the  external  resistances.  These  curves 
are  found  to  coincide  very  closely  with  the  curves  obtained  from 
the  brake  test,  provided  that  the  brake  power  and  the  fir  power 
are  plotted  to  the  same  scale,  one  representing  the  mechanical, 
the  other  the  corresponding  electrical  output.  Some  difference 
in  the  curves  is  due  to  the  fact  that  the  stationary  transformer 
has  no  friction  loss;  this  is,  however,  partly  or  wholly  compen- 
sated by  a  greatly  increased  secondary  core  loss. 

The  theoretical  reasons  for  this  equivalence  of  an  induction 
motor  to  a  polyphase  transformer  will  become  clear  by  consider- 

1  See,  for  instance,  the  author's  Experimental  Electrical  Engineering,  Vol.  1, 
chap.  17. 

122 


CHAP.  XII]  THE  INDUCTION   MOTOR  123 

ing  the  electrical  relations  in  the  rotor  under  the  following  three 
headings: 

(a)  The  Relationship  between  the  External  Resistance  and  the 
Slip.     Let  the  input  into  the  rotor  be  P  watts  per  phase  of  the 
secondary  winding,  and  let  the  motor  be  running  at  a  slip  s.     For 
instance,  s  =  0.05  means  that  the  speed  of  the  rotor  is  5  per  cent 
lower  than  the  synchronous  speed,  or  the  speed  of  the  revolving 
field.     Then,  sP  watts  are  converted  into  heat  in  each  phase  of 
the  secondary  winding,  and  (1  —  s)P  watts  are  available  on  the 
shaft  as  the  output  (including  friction  and  windage).     This  is 
because  the  tangential  electromagnetic  effort  is  the  same  on  the 
surface  of  the  stator  as  it  is  on  the  surface  of  the  rotor.    But 
while  the  gliding  magnetic  flux  travels  at  synchronous  speed,  the 
rotor  travels  at  (1  —  s)  times  the  synchronous  speed.     The  elec- 
tromagnetic coupling  between  the  stator  and  the  rotor  is  simi- 
lar to  a  friction  coupling  between  two  shafts,  having  a  certain 
amount  of  slip.     If  the  speed  of  the  driven  shaft  is  say  5  per  cent 
below  that  of  the  driving  shaft,  on  account  of  the  slip  in  the 
coupling,  95  per  cent  of  the  power  is  transmitted  and  5  per  cent 
is  lost  in  heat  in  the  coupling. 

With  the  rotor  blocked,  let  R  be  the  external  resistance  per 
phase  of  the  secondary,  and  let  J2  be  the  secondary  current  per 
phase.  For  a  slip  s  we  must  have  the  condition 

s  (R  +  r2)  722  =  r2/22, 
or 

s  =  r2/(fl  +  r2) (206) 

If  the  slip  is  given,  the  required  external  resistance  is 

R  =  ra(l  -  s)/s (207) 

(b)  Equal  Secondary  Current  and  Phase  Displacement  with  the 
Rotor  Running  or  Blocked.     Let  the  reactance  of  the  secondary 
winding  per  phase  be  Xz  ohms,  at  the  primary  or  synchronous 
frequency.     With  the  rotor  running  at  a  slip  s,  the  frequency  of 
the  secondary  currents  is  only  equal  to  s  times  the  primary  fre- 
quency, so  that  the  reactance  per  phase  is  sx^.     Therefore,  the 
phase  displacement  02  between  the  induced  secondary  voltage 
and  the  current  is  determined  by  the  relation 

tan  02  =  sz2/r2 (208) 

With  the  rotor  blocked  and  provided  with  external  resistances 
satisfying  condition  (206),  the  total  resistance  of  the  secondary 


124  THE  ELECTRIC  CIRCUIT  [ART.  42 

circuit  per  phase  is  R  +  r2  =  ra/s.  The  secondary  frequency 
is  equal  to  that  in  the  primary  circuit,  so  that 

tan  <£2  =  Xt/(r2/s}  =  sx2/rz; 

thus  the  phase  displacement  is  the  same  as  that  given  by  eq.  (208). 
With  the  same  revolving  magnetic  flux  in  both  cases,  the  currents 
are  also  equal.  While  with  the  stationary  rotor  the  induced 
secondary  e.m.f.  is  larger  in  the  ratio  of  1:  s,  because  of  a  higher 
speed  of  cutting  the  secondary  conductors,  yet  the  total  secondary 
resistance  R  +  r2  is  also  larger  in  the  same  ratio  of  1 :  s,  according 
to  cq.  (206).  The  secondary  reactance  is  also  larger  in  the  same 
ratio  on  account  of  the  higher  frequency.  Thus,  with  the  rotor 
blocked,  both  the  e.m.f.  and  the  impedance  of  the  secondary 
circuit  are  larger  in  the  ratio  of  1 :  s  than  when  it  is  running  at  a 
slip  s.  Hence,  the  current,  which  is  equal  to  the  ratio  of  the  e.m.f. 
to  the  impedance,  is  the  same  in  both  cases. 

(c)  The  Reaction  of  the  Secondary  upon  the  Primary  Circuit 
is  the  Same  with  the  Rotor  Running  or  Blocked.  The  magneto- 
motive force  of  the  revolving  rotor  is  the  same  as  that  of  the 
stationary  rotor  with  the  resistance  R  in  series,  provided  that  in 
both  cases  the  magnetomotive  force  is  considered  with  respect  to 
the  stationary  primary  circuit.  In  the  latter  case  the  frequency 
of  the  secondary  currents  is  equal  to  that  of  the  supply,  so  that 
the  resultant  magnetomotive  force  due  to  all  the  secondary  phases 
travels  in  the  air-gap  at  synchronous  speed,  the  same  as  the 
resultant  magnetomotive  force  of  the  primary  currents.  The 
two  magnetomotive  forces  ^form  one  resultant  magnetomotive 
force  which  produces  the  revolving  flux.  With  the  revolving 
rotor,  the  frequency  of  the  secondary  currents  is  s  per  cent  of  that 
of  the  supply,  so  that  the  secondary  magnetomotive  force  glides 
relatively  to  the  body  of  the  rotor  at  a  speed  equal  to  s  per  cent 
of  the  synchronous  speed.  But  the  speed  of  the  rotor  itself  is 
the  (1  —  s)  part  of  the  synchronous  speed.  Hence,  the  velocity 
of  the  secondary  magnetomotive  force  with  respect  to  the  stator  is 
s  +  (1  ~  s)  =  1,  or  is  equal  to  the  synchronous  speed,  and  is 
the  same  as  the  velocity  of  the  primary  magnetomotive  force. 
We  have  seen  above  that  the  secondary  currents  and  their  phase 
relation  are  the  same  in  the  two  cases,  so  that  the  secondary 
magnetomotive  force  is  also  the  same  in  phase  and  magnitude. 
Consequently,  with  the  same  flux,  determined  by  the  applied  volt- 


CHAP.  XII]  THE  INDUCTION  MOTOR  125 

age,  the  primary  magnetomotive  force  is  also  the  same  in  both 
cases.  This  means  that  the  primary  current  and  power-factor 
are  the  same  with  the  stationary  rotor  loaded  electrically  as  with 
the  revolving  rotor  loaded  mechanically. 

We  have  thus  proved  theoretically,  as  well  as  experimentally, 
that  the  performance  of  an  induction  motor  may  be  reduced  to 
that  of  a  stationary  transformer.  But  we  know  from  the  pre- 
ceding chapter  that  a  transformer  can  in  turn  be  replaced  by  an 
equivalent  electric  circuit,  either  approximately  (Fig.  41)  or  accu- 
rately (Fig.  42).  Thus,  the  same  equivalent  diagrams  can  be  used 
in  the  predetermination  of  the  performance  of  an  induction  motor. 
All  the  quantities  which  enter  into  these  diagrams  are  understood 
to  be  per  phase  of  the  primary  circuit  (usually  per  phase  of  Y 
in  a  three-phase  motor).  When  the  number  of  the  secondary 
phases  and  the  method  of  connections  are  different  from  those 
in  the  primary  circuit,  the  secondary  winding  is  replaced  by  an 
equivalent  one  of  the  same  number  of  phases,  and  with  the  same 
kind  of  connections  as  in  the  primary  circuit;  see  Art.  45  below.1 

Prob.  1.  Explain  the  principle  of  the  speed  control  of  an  induction 
motor  by  means  of  adjustable  external  resistances  in  the  secondary  cir- 
cuit. 

Prob.  2.  Explain  the  principle  of  the  direct  and  differential  cascade 
connection  of  two  induction  motors. 

Prob.  3.  Show  how  an  induction  generator  can  be  reduced  to  an 
equivalent  electric  circuit. 

43.  The  Analytical  Determination  of  Performance.  —  Ap- 
proximate Solution.  The  problem  is  to  calculate  the  perform- 
ance characteristics  of  a  given  induction  motor — in  other  words, 
against  the  output  as  abscissae,  to  plot  the  following  curves;  viz., 
primary  amperes,  kilowatts  input,  primary  power-factor,' slip, 
torque,  and  efficiency.  The  resistances  and  the  leakage  reactances 
of  both  windings,  reduced  to  the  primary  circuit,  are  supposed  to 
be  known,  so  that  each  primary  phase  of  the  motor  can  be  replaced 
by  either  the  approximate  or  the  exact  diagram  (Figs.  41  and  42). 
The  approximate  diagram  only  is  considered  here,  because  it  is 
sufficiently  accurate  for  most  practical  purposes.  The  exact  solu- 
tion is  given  in  Art.  47  below.  The  iron  loss,  friction,  and  the 

1  The  values  of  the  leakage  reactances  of  the  windings  are  supposed  here 
to  be  known;  for  their  calculation  from  the  dimensions  of  the  motor  see  the 
author's  Magnetic  Circuit,  Art.  66. 


126  THE  ELECTRIC  CIRCUIT  [ART.  43 

magnetizing  current  are  also  supposed  to  be  known,  so  that  the 
exciting  admittance  Y0  is  known.  While  in  reality  it  varies  some- 
what  with  the  load,  in  the  approximate  solution  it  is  considered 
to  be  a  constant  quantity. 

The  problem  is  solved  similarly  to  that  of  the  voltage  regula- 
tion of  a  transmission  line  or  of  a  transformer,  treated  above; 
that  is,  a  load  current  IL  is  selected,  and  the  circuit  is  solved  in 
the  complex  notation.  In  order  to  make  the  treatment  independ- 
ent of  the  other  chapters,  a  complete  solution  is  given  below,  with 
some  minor  changes  which  simplify  the  numerical  work.  As  in 
the  transmission  line  and  in  the  transformer,  we  have 

Ei=*EL  +  ZILt (209) 

or,  expanded, 

Ei  (cos  0  +  j  sin  0)  =  EL  +  IL  (r  +  jx) .  .  .  (210) 
Here  the  direction  of  the  unknown  load  voltage  EL  is  again  selected 
as  the  reference  axis.  The  current  IL  is  in  phase  with  EL,  because 
by  assumption  the  external  resistance  R  is  non-inductive.  Sepa- 
rating the  real  and  the  imaginary  parts,  we  get 

Ei  cos  9  =  EL  +  ILr; (211) 

Ei  sin  6  =  ILx (212) 

When  plotting  the  curves,  it  is  immaterial  which  values  of  the  load 
are  selected  for  computation.  We  assume,  therefore,  a  series  of 
reasonable  values  for  IL,  and  from  eq.  (212)  calculate  the  corre- 
sponding values  of  sin  6.  Then  from  eq.  (211)  we  find  the  values 
of  EL,  and  finally  determine  the  outputs  per  phase  from  the 
equation 

PL  =  !LEL (213) 

Knowing  IL,  EL,  and  the  angle  6,  the  rest  of  the  values  for  the 
performance  curves  are  calculated  as  follows: 

(a)  The  Slip.     The  external  resistance,  reduced  to  the  primary 
circuit,  is 

R'  =  EL/IL- (214) 

and  the  slip  is  found  from  the  equation 

s  =  iV/GR'  +  r,'), (214a) 

which  is  identical  with  eq.  (206),  except  that  r2'  and  R'  are  second- 
ary quantities  reduced  to  the  primary  circuit. 

(b)  The  Primary  Current  and  Power-factor.    The  total  primary 
current  per  phase  is 

!i=  !*+!L, (215) 


CHAP.  XII]  THE  INDUCTION  MOTOR  127 

where  the  magnetizing  current  /0  is  known,  and  can  be  repre- 
sented with  respect  to  the  terminal  voltage  E:  as 

/o  =  /0  (cos  </>o  —  jsin^o)  .....  (216) 
The  load  current,  in  its  phase  relation  with  respect  to  the  terminal 
voltage,  is 

IL  =  IL(cos9-  jsinfl),  .....     (217) 
so  that 

!i  =  ii-jii  =  (!L  cos  0+/o  cos  00)  —J(!L  sin  0+/o  sin  00).    (218) 
Knowing  the  projections  ii  and  i\   of  the  primary  current  with 
respect  to  the  terminal  voltage  E\,  the  primary  phase  angle  0i 
is  found  from  the  equation 

tan  0i  =  *YAi,      ......     (219) 

and  then  the  primary  power-factor,  cos  0i,  is  taken  from  a  trigo- 
nometric table.     The  current  itself, 

7i  =  ii/cos0i  .......     (220) 

(c)  The  input  per  phase  is 

Pi  =  Eil  i  cos  0i  =  Eiii  .....     (221) 
The  efficiency  is  equal  to  the  ratio  of  the  output  to  the  input. 

(d)  The  useful  torque  in  synchronous  watts,  or  the  input  into 
the  secondary,  is  equal  to  the  output  plus  the  secondary  copper 
loss.     The  tangential  effort  per  phase,  in  kilograms  at  a  radius  of 
one  meter,  or  the  torque  per  phase,  in  kg.-meters  is 

T  =  973.8  (PL  +  0.001  7LV2')/(synchr.  r.p.m.),  .  (222) 
PL  being  expressed  in  kilowatts,  so  as  to  avoid  large  numbers  in 
numerical  applications. 

Sometimes  the  performance  data  are  desired  for  one  particular 
load,  PL,  only;  for  instance,  at  the  rated  output  of  the  machine. 
The  method  outlined  above  may  in  this  case  be  somewhat  tedious, 
because  one  has  to  find  by  trials  the  proper  values  of  IL  and  EL 
which  give  the  desired  output.  It  may  lead  more  quickly  to  the 
desired  end  to  solve  eqs.  (211),  (212),  and  (213)  as  three  simul- 
taneous equations  for  the  unknown  quantities  EL,  IL,  and  6. 
Squaring  the  first  two  equations  and  adding  them  together,  the 
angle  6  is  eliminated,  and  we  get 


Substituting  for  EL  its  value  from  eq.  (213),  gives  a  quadratic 
equation  for  z/L2,  namely 

(zIL*Y  -2  (z7L2)  (i  tfi2  -  PLr}/z  +  PL*  =  0.  .     .     (223) 


^4.       CVt 


128  THE  ELECTRIC  CIRCUIT  [ART.  43 

The  solution  of  this  equation  is 


2/L2  =  (i  #i2  -  PiA/z  -      (l  Ei*  -Pir)*/*  -Pi*. 
The  minus  sign  only  is  retained  before  the  radical,  because  it 
gives  a  smaller  current.     It  can  be  shown  that  the  solution  with 
the  plus  sign  corresponds  to  the  unstable  region  of  operation  of 
the  motor. 

For  numerical  computations  the  preceding  equation  is  put  in 
the  form 


zlj}  =  1000  (Q  -  VQ2  _  pL2),      .     .     .     (224) 
where,  for  the  sake  of  brevity,  we  introduce  the  notation, 

Q=  (5QOES- PLr)/z (225) 

In  the  last  two  equations  PL  and  Q  are  in  kilowatts,  so  as  to  avoid 
large  numbers,  and  EI  is  in  kilovolts.  The  student  is  reminded 
that  EI  is  the  phase  or  star  voltage,  and  not  the  line  voltage,  and 
that  PL  is  the  output  per  phase. 

When  PL  is  small  compared  to  Q,  formula  (224)  represents 
the  difference  of  two  quantities  of  nearly  equal  value.  The 
result  is  inaccurate,  and  it  is  better  to  expand  the  expression 
[1  —  (Pz,/Q)2]*>  according  to  the  binomial  theorem.  This  gives 

zlj}  =  1000  Q  ft  (PL/QY  +  i  (PL/QY  +  ^  (PL/QY  +  etc.].  (226) 

The  latter  formula  is  much  more  convenient  for  numerical  applica- 
tions than  eq.  (224),  because  the  second  term  in  the  brackets  is 
small  as  compared  to  the  first,  and  the  third  term  can  usually  be 
neglected. 

Knowing  IL,  the  rest  of  the  values  are  determined  as  before. 

Prob.  1.  Plot  complete  performance  curves  of  a  three-phase,  25- 
cycle,  150-kw.,  6-pole,  2200-volt,  induction  motor  between  no  load  and 
25  per  cent  overload,  from  the  following  data:  Total  no-load  input  (for 
all  three  phases)  is  10.5  kw.;  the  no-load  current  per  phase  of  the  line 
is  13  amp.  With  the  armature  blocked,  the  input  is  230  kw.,  the  cur- 
rent per  phase  being  227  amp.1  The  resistance  of  the  primary  winding 
per  phase  of  Y  is  0.60  ohm.  Hint:  Follow  consistently  the  approximate 
diagram,  Fig.  41;  that  is,  do  not  correct  the  no-load  reading  for  the 
primary  i-r  loss,  and  assume  the  magnetizing  current  with  the  armature 
locked  to  be  the  same  as  at  no  load. 

1  The  data  with  the  armature  locked  refer  to  the  rated  voltage;  they 
are  obtained  by  extrapolating  the  curves  taken  at  lower  voltages.  It  would 
not  be  practicable  to  apply  the  full  line  voltage  to  a  large  motor  with  the 
armature  blocked. 


CHAP.  XII]  THE  INDUCTION  MOTOR  129 

ABS.  At  the  rated  output  the  primary  current  is  49  amp.;  the 
power-factor  is  90.8  per  cent;  the  slip,  3.5  per  cent;  the  efficiency,  88.5 
per  cent;  and  the  torque,  302  kg.-m. 

Prob.  2.    Check  the  answer  to  the  preceding  problem,  using  eq.  (226). 

Prob.  3.  Extend  the  theory  and  the  formulae  given  above  to  the 
performance  characteristics  of  an  induction  generator. 

44.  Starting  Torque,  Pull-out  Torque,  and  Maximum  Out- 
put. When  judging  the  performance  of  a  given  induction 
motor,  or  designing  a  new  motor,  the  following  features  are  of 
importance : 

(a)  The  starting  torque,  either  in  its  absolute  value,  or  in 
its  ratio  to  the  torque  at  the  rated  load.     If  the  motor  is  to  be 
started  by  means  of  resistances  in  the  secondary  circuit,  one  may 
be  required  to. calculate  the  values  of  these  resistances  necessary 
for  a  prescribed  starting  torque,   or  for  a  maximum  starting 
torque. 

(b)  The  pull-out  torque,  or  the  torque  at  which  the  motor 
reaches  the  limit  of  stable  operation,  and  comes  to  a  stop.     This 
torque  is  usually  given  through  its  ratio  to  the  full-load  torque. 

(c)  The  maximum  output  of  the  motor,  in  kilowatts.     This 
output  takes  place  at  a  smaller  slip  than  that  at  which  the  motor 
pulls  out.     The  output  is  a  maximum  when  the  product   of 
torque  times  speed  is  a  maximum,  but  not  when  the   torque 
itself  is  greatest. 

The  three^  quantities  mentioned  above  can  be  determined  by 
using  the  equations  deduced  in  the  foregoing  article. 

(a)  The  Starting  Torque.  In  the  general  formula  (222),  PL  =  0 
at  start,  because  the  motor  supplies  no  mechanical  output,  the 
speed  being  equal  to  zero.  In  the  equivalent  electrical  diagram 
(Fig.  41)  this  corresponds  to  a  short  curcuit  of  the  load,  or  R  =  0. 
Hence,  IL  =  E\/z;  substituting  this  value  into  eq.  (222),  we  find 
that  the  starting  torque  per  phase,  in  kg.-  m. 

T8t  =  0.9738  #i2r2'/ (z2  X  synchr.  r.p.m.).  (227) 

It  will  be  seen  from  this  expression  that  the  starting  torque  is 
proportional  to  the  square  of  the  line  voltage.  This  fact  permits 
one  to  determine  the  initial  torque  when  starting  a  motor  on  a 
lower  voltage,  by  means  of  auto-transformers.  The  same  equa- 
tion shows  that  the  starting  torque  increases  with  the  secondary 
resistance.  It  is  not  quite  proportional  to  it,  because  r2'  is  also 
implicitly  contained  in  z. 


130  THE  ELECTRIC  CIRCUIT  [ART.  44 

When  the  motor  has  a  phase-wound  secondary  and  is  started 
by  means  of  resistances  in  the  secondary  circuit,  r*  in  formula 
(227)  includes  this  secondary  resistance.  It  is  thus  possible  to 
calculate  the  external  resistance  required  for  a  given  starting 
torque.  For  example,  when  the  starting  torque  is  given,  the 
ratio  z2/r2'  in  eq.  (227)  is  a  known  quantity,  or 

*/rt'  =  [(n  +  r/)2  +  *W  =  c,      .     .     .     (228) 
where 

c  =  0.9738  #i2/ (desired  torque  X  synchr.  r.p.m.).       (229) 

Solving  the  quadratic  (228)  for  r/,  we  obtain 

*).  (230) 


In  applications,  the  minus  sign  only  is  retained  before  the  radical 
because  one  would  naturally  use  the  smaller  of  the  two  resistances 
which  give  the  same  torque. 

The  value  of  r2'  determined  from  this  expression  comprises 
both  the  resistance  per  phase  of  the  rotor  proper  and  the  starting 
resistance  per  phase,  if  any  is  used,  both  reduced  to  equivalent 
primary  values.  To  obtain  their  actual  values,  see  Art.  45  below. 

If  the  external  resistance  is  to  be  so  selected  as  to  give  a  max- 
imum starting  torque,  c  in  expression  (228)  must  be  a  minimum. 
Equating  to  zero  the  derivative  of  c  with  respect  to  r/,  and 
solving  for  r2',  we  get 

r,'  =  (z2  +  ri»)*, (231) 

that  is,  r-i  is  very  nearly  equal  to  x.  This  value  comprises  the 
resistance  of  the  rotor  proper  and  the  starting  resistance,  both 
per  phase  of  the  primary  circuit.  If  a  resistance  is  selected  which 
is  either  less  than  or  greater  than  that  determined  by  eq.  (231), 
the  motor  does  not  develop  its  full  starting  torque.  This  checks 
with  eq.  (230),  which  shows  that  the  same  torque  can  be  obtained 
with  two  different  values  of  starting  resistance. 

(b)  Pull-out  Torque.  According  to  eq.  (222),  the  torque  is  a 
maximum  when 

EJL  +  IM  =  max (232) 

Here  EL  and  IL  are  functions  of  the  independent  variable  6. 
Expressing  them  through  0  from  eqs.  (211)  and  (212),  and  omit- 
ting the  constant  factor  #i2,  we  obtain 

(1/z)  sin  9  [cos  0  -  (r/z)  sin  0]  +  (r2'/z2)  sin2  0  =  max., 


CHAP.  XII]  THE  INDUCTION  MOTOR  131 

or,  after  simplification, 

x  sin  2  0  -  ri  (1  -  cos  26)  =  max.     .     .     .     (233) 

Equating  to  zero  the  derivative  of  this  expression  with  respect  to 
6,  gives 

x  cos  2  6  -  ri  sin  2  6  =  0, 
or 

tan  2  6  =  x/ri (234) 

Knowing  6,  the  values  of  EL  and  IL  are  calculated  from  eqs.  (211) 
and  (212),  and  then  the  torque  is  determined  from  eq.  (222). 

It  is  of  interest  to  note  that  the  angle  0,  at  which  the  motor 
pulls  out  of  step,  is  independent  of  the  secondary  resistance 
r2'.  Neither  does  this  resistance  enter  into  eq.  (233).  Hence, 
the  maximum  torque  which  a  motor  is  capable  of  developing  is  inde- 
pendent of  its  rotor  resistance.  This  resistance  determines  only  the 
speed  at  which  the  maximum  torque  takes  place.  The  higher  the 
secondary  resistance,  the  lower  the  speed  at  which  the  motor 
pulls  out  of  step.  By  using  an  external  starting  resistance,  and 
a  rotor  winding  of  low  resistance,  two  maxima  of  the  torque  are 
obtained,  one  at  the  start,  with  the  external  resistance  in,  and  the 
other  near  synchronism,  with  it  out. 

(c)  Maximum  Output.  The  problem  is  to  find  the  values  of 
EL  and  IL  for  which  the  product  EL!L  is  a  maximum.  Again 
expressing  EL  and  IL  through  the  angle  6  from  eqs.  (211)  and  (212), 
and  omitting  the  constant  factor  Ei*/x,  we  have 

sin  6  [cos  0  —  (r/x)  sin  6}  =  max. 

Equating  to  zero  the  derivative  of  this  expression  with  respect  to 
0,  gives 

*  =  **, (235) 

where  tan  <f>  =  x/r.  Knowing  the  angle  0,  the  values  of  IL  and 
EL  are  calculated  from  eqs.  (211)  and  (212),  and  then  their  product 
EL!L  is  determined. 

Prob.  1.  The  motor  specified  in  problem  1  of  the  preceding  article 
is  designed  to  be  started  at  a  reduced  voltage.  What  per  cent  tap  should 
be  used  on  the  auto-transformers  in  order  to  get  a  starting  torque  of 
about  30  per  cent  of  the  full-load  torque? 

Ans.     60  per  cent  of  the  line  voltage. 

Prob.  2.  The  same  motor  is  provided  with  a  phase-wound  secondary 
and  is  to  be  started  by  using  resistances  in  series  with  the  rotor  windings. 


132  THE  ELECTRIC  CIRCUIT  [AaT.  44 

What  external  resistance  is  necessary  in  order  to  obtain  a  starting  torque 
equal  to  1.5  times  the  full-load  torque? 

Ans.     0.78  ohm  per  phase,  in  terms  of  the  primary  circuit. 

Prob.  3.  What  starting  resistance  in  the  preceding  problem  would 
give  the  maximum  starting  torque?  Ans.  About  5.7  ohms. 

Prob.  4.  Show  that  the  motor  specified  in  problem  1  of  the  preceding 
article  pulls  out  of  step  when  the  torque  exceeds  2.45  times  the  rated 
full-load  torque. 

Prob.  5.  Check  the  answer  to  problem  4  by  using  the  answer  to  prob- 
lem 3. 

Prob.  6.  Show  that  the  maximum  output  of  the  same  motor  is  equal 
to  2.15  times  the  rated  output. 

Prob.  7.  Show  that  the  input  into  an  induction  motor  is  a  maximum 
when  6  =  45°.  Hint:  IL  cos  6  =  max. 

Prob.  8.  Show  how  to  calculate  the  per  cent  slip  at  which  the  motor 
pulls  out  of  step,  and  also  the  speed  at  which  the  output  is  a  maximum. 


CHAPTER  Xin 

PERFORMANCE  CHARACTERISTICS  OF  THE 
INDUCTION  MOTOR—  (Continued) 

45.  The  Secondary  Resistances  and  Reactances  Reduced 
to  the  Primary  Circuit.  It  is  proved  in  Art.  40  that  in  a  trans- 
former the  secondary  resistance  and  reactance  can  be  transferred 
into  the  primary  circuit  by  being  multiplied  by  (rii/w2)2.  The 
same  rule  holds  true  for  the  induction  motor,  provided  that  the 
number  of  phases  is  the  same  in  the  primary  and  in  the  secondary 
windings,  and  that  the  two  windings  are  of  the  same  type  (the 
same  number  of  slots  per  phase  and  the  same  winding  pitch). 
This  is  hardly  ever  the  case,  and  with  a  different  number  of 
phases  and  different  types  of  winding  in  the  primary  and  second- 
ary, the  following  formula  holds  true : 

r2'/r2  =  (roi/rog)  (fc6in1//c62n2)2,  ....     (236) 
and  analogously  for  the  reactances, 

x2'/xz  =  (mi/m2)  (fcmi/fan,)*.      .     .     .     (237) 

In  these  expressions,  m  stands  for  the  number  of  phases,  n  is  the 
number  of  turns  per  phase,  and  fa  is  the  so-called  breadth  factor 
which  characterizes  the  winding.  The  subscripts  1  and  2  refer 
to  the  primary  and  secondary  windings  respectively.  The  quan- 
tities r2  and  z2  are  the  actual  resistance  and  reactance  per  phase 
of  the  secondary  circuit;  r/  and  Xzf  are  the  equivalent  quantities 
per  phase  of  the  primary  circuit.1  When  Wi  =  ra2  and  kbi  =  fc&2, 
the  preceding  formulae  become  identical  with  eqs.  (196)  and  (196a) 
for  the  transformer. 

Equations  (236)  and  (237)  refer  to  the  resistances  and  react- 
ances per  phase,  with  the  understanding  that  the  windings  of 
each  phase  are  all  in  series,  both  in  the  stator  and  in  the  rotor; 
and  that  the  connections  are  either  both  star,  or  both  mesh,  even 
if  the  number  of  phases  be  different.  Otherwise,  the  actual  con- 

1  For  a  proof  of  these  formulae,  see  the  author's  Magnetic  Circuit,  Art.  44; 
the  values  of  kb  will  be  found  in  Arts.  27  to  29  of  the  same  book. 

133 


134 


THE   ELECTRIC  CIRCUIT 


[AKT.  45 


ncctions  must  be  taken  into  consideration  and  the  values  of  r2' 
and  x2'  further  modified,  keeping  in  mind  the  fact  that  the  total 
tV  loss  must  be  the  same  in  the  equivalent  winding  as  in  the  ac- 
tual one.  As  a  simple  illustration,  let  the  stator  be  three- 
phase  Y-connected,  and  the  rotor  three-phase  delta-connected.  If 
the  equivalent  secondary  resistance  calculated  by  means  of 
eq.  (236)  is  r/,  then  only  ^  r2'  must  be  used  in  the  equivalent 
diagram,  per  phase  of  Y.  This  is  because  the  current  per  phase 
of  Y  is  v3  times  as  large  as  that  per  phase  of  delta,  hence,  for 
the  same  &r  loss,  the  resistance  per  phase  of  Y  must  be  only  one- 
third  of  that  per  phase  of  delta.  The  same  relation  holds  true 
for  reactances,  because  the  stored  electromagnetic  energy  is  also 


FIG.  43.  A  squirrel-cage  rotor  and  the  FIG.  44.  The  vectorial  relation  between 
star  resistances  equivalent  to  the  end  the  star  and  mesh  currents  in  a  sym- 
ring.  metrical  m-phase  system. 

proportional  to  the  square  of  the  current  (Art.  20),  and  the 
equivalent  diagram  must  express  correctly  the  cyclic  exchange 
of  energy  between  the  primary  and  the  secondary  circuits. 

In  the  most  general  case,  let  an  ra-phase  symmetrical  system 
be  given,  for  instance  a  two-pole  squirrel-cage  rotor  (Fig.  43), 
and  let  it  be  required  to  find  the  relation  between  the  mesh  resist- 
ances rm  and  the  star  resistances  r,  such  that  the  izr  loss  per  phase 
shall  be  the  same  in  both.  The  relation  between  the  vectors  of 
the  star  currents  and  those  of  the  mesh  currents  is  shown  in  Fig.  44, 
the  star  currents  /,  forming  an  m-sided  polygon,  and  the  mesh 
currents  Im  being  the  radii  of  the  polygon.  This  diagram  is  correct 
because  it  satisfies  the  following  conditions :  (a)  the  star  currents 


CHAP.  XIII]  THE  INDUCTION   MOTOR  135 

are  displaced  in  phase  relatively  to  each  other  by  equal  angles 
of  2  TT/W,  over  the  whole  range  of  2  TT;  (b)  the  same  is  true  for 
the  mesh  currents;  (c)  each  star  current  is  the  geometric  differ- 
ence of  the  two  adjacent  mesh  currents;  (d)  the  geometric  sum  of 
the  star  currents  is  equal  to  zero  (Kirchhoff's  first  law).  From 
the  geometry  of  the  figure  we  have 

H,  =  /msin(7r/m)  ......     (238) 

The  condition  782rg  =  7m2rm  leads  to  the  ratio 

m).      .     .     .     (230) 


A  similar  relation  holds  for  the  reactances.     When  m  =  3,  we 
find  as  before  that  rA/ry  =  3. 

Equation  (239)  finds  its  practical  application  in  the  calcula- 
tion of  the  equivalent  resistance  and  reactance  of  a  squirrel-cage 
rotor.  The  sections  of  the  end-rings  between  the  bars  are  mesh- 
connected,  while  the  bars  themselves  may  be  considered  as  parts 
of  a  star-connected  w2-phase  system,  where  w2  is  the  number  of 
bars  per  pair  of  poles.  Let  r&  be  the  resistance  of  each  bar,  includ- 
ing the  two  contact  resistances  between  the  end-rings  and  the 
bar;  let  rr  be  the  resistance  of  a  section  of  an  end-ring  between 
two  consecutive  bars.  The  resistance  of  the  rings  can  be  replaced 
by  added  resistances  in  series  with  the  bars,  so  as  to  change  the 
connections  to  a  pure  star  system  (Fig.  43).  According  to  eq. 
(239)  we  find  that  the  new  resistance  per  bar  must  be  equal  to 
r&  +  2  rr/[4  sin2  (7r/m2)].  If  the  motor  has  p  poles,  or  \  p  pairs  of 
poles,  there  are  \  p  bars  in  parallel  belonging  to  the  same  phase,  so 
that  the  total  resistance  of  the  secondary  winding  per  phase  is  only 
the  2/-p  part  of  that  of  one  bar.  Hence,  assuming  that  the  primary 
winding  is  star-connected,  and  that  all  the  coils  in  each  phase  are 
in  series,  the  value  of  r2  to  be  used  in  eq.  (236)  is 


.     .     .     (240) 
Analogously, 

.     .     .     (241) 


Since  there  is  only  one  bar  per  phase,  and  one  bar  is  equivalent 
to  one-half  of  a  turn,  the  value  n2  =  ^  and  fc&2  =  1  must  be 
used  hi  eqs.  (236)  and  (237). 

Prob.  1.  A  two-phase  induction  motor  has  the  primary  winding  ar- 
ranged for  two  independent  phases;  the  secondary  is  three-phase  Y-con- 


136 


THE  ELECTRIC  CIRCUIT 


[AiiT.  46 


ncctcd.  When  the  rotor  is  stationary  and  its  circuits  arc  open,  440  volts 
impressed  at  the  primary  terminals  produce  97  volts  between  the  slip- 
rings.  The  calculated  starting  resistance,  per  phase  of  the  primary 
circuit,  is  14  ohms.  What  is  the  actual  resistance  to  be  used  in  series  with 
the  rotor  windings?  Hint:  Consider  the  primary  circuit  as  ajour-phase 
star-connected  system,  so  that  njtbi/njtbz  =  I  X  440/(97/  V3). 

Ans.     0.34  ohm. 

Prob.  2.  A  six-pole,  three-phase,  Y-connected  induction  motor  has  a 
squirrel-cage  rotor  of  80  cm.  diameter  with  73  bars;  the  resistance  of 
each  bar  is  120  microhms  (including  the  contact  resistance).  In  order 
to  have  a  certain  required  torque  and  slip,  the  equivalent  rotor  resistance 
per  phase  of  the  primary  circuit  must  be  equal  to  1.07  ohms.  What  must 
be  the  actual  resistance  of  each  end-ring  per  centimeter  of  its  length? 
There  are  100  turns  per  phase  of  the  primary  winding,  and  kb\  =  0.95; 
for  the  squirrel-cage  winding  kb2  is  always  equal  to  unity. 

Ans.     5.8  microhms. 

46.  The  Circle  Diagram.  Let  the  values  of  primary  current 
obtained  from  a  brake  test  on  an  induction  motor  be  plotted  as 
vectors  at  proper  phase  angles  with  respect  to  the  vector  EI 
of  the  primary  voltage  (Fig.  45).  The  locus  of  the  ends  of  the 
current  vectors  is  found  to  be  very  nearly  a  semicircle.  This 


i?  i;/ 

FIG.  45.     The  circle  diagram  of  an  induction  motor. 


locus,  together  with  some  auxiliary  lines,  is  called  the  circle  dia- 
gram or  the  Heyland  diagram  of  the  induction  motor.  A  similar 
diagram  holds  true  for  the  transformer,  although  it  is  hardly  ever 
used  in  practice. 

The  importance  and  the  convenience  of  the  circle  diagram  lie 
in  the  fact  that  the  complete  performance  of  an  induction  motor 
can  be  predicted  if  the  diameter  of  the  semicircle  and  its  posi- 


CHAP.  XIII]  THE  INDUCTION   MOTOR  137 

tion  with  respect  to  the  voltage  vector  EI  are  known.  The  semi- 
circle is  usually  determined  by  the  vector  70  of  the  no-load  current, 
and  the  vector  of  the  current  I8  obtained  when  the  rotor  is  locked 
(the  subscript  s  stands  for  starting  or  short  circuit).  At  any 
other  load,  the  extremity  of  the  current  vector  7i  lies  between 
those  of  70  and  78. 

The  Heyland  diagram  is  simply  a  graphic  representation  of 
the  current  and  voltage  relations  in  the  approximately  equivalent 
circuit  diagram  shown  in  Fig.  41.  The  exciting  current  and  the 
no-load  losses  are  assumed  to  be  constant  at  all  loads  from  no- 
load  to  standstill.  The  no-load  current  is  resolved  into  a  loss 
component  /</',  in  phase  with  the  voltage  E\,  which  component 
represents  the  iron  loss,  friction,  and  windage;  and  a  reactive  com- 
ponent Id  which  excites  the  mam  flux  in  the  motor.  At  any  load, 
the  primary  current  I\  is  the  geometric  sum  of  the  load  current 
IL  and  the  no-load  current  70,  the  vectors  of  these  three  currents 
forming  a  triangle. 

That  the  locus  of  the  current  7i  or  IL  is  a  circle  follows  directly 
from  eq.  (212),  because  from  it  we  have 

/L/sin  e  =  Ei/x  =  const.,      ....     (242) 

which  is  easily  seen  to  be  the  equation  of  a  circle  in  polar  coordi- 
nates. The  value  of  the  constant 

Edx  =  I  if (243) 

is  equal  to  the  diameter  of  the  circle,  which  diameter  is  thus 
determined  solely  by  the  leakage  reactance  x  of  the  motor.  The 
smaller  x  is,  the  larger  is  the  circle  and  the  better  the  motor, 
because  its  power-factor  is  higher  and  its  overload  capacity 
larger. 

The  load  current  I,L  with  the  armature  blocked  has  an  energy 
component  in  phase  with  the  line  voltage  E\,  because  of  the  i?r 
loss  in  the  resistances  of  the  stator  and  rotor.  If  these  resistances 
could  be  eliminated  or  put  outside  the  motor,  the  load  current 
on  short  circuit  would  be  purely  reactive  and  equal  to  IL'  =  EI/X. 
Thus,  the  diameter  of  the  circle  is  equal  in  position  and  magni- 
tude to  the  load  current  (secondary  current)  of  the  machine  with 
the  armature  blocked,  provided  that  the  internal  resistances  are 
eliminated  and  only  leakage  reactances  are  left.  This  condition 
is  called  an  ideal  short  circuit. 

Leaving  the  loss  component  70"  of  the  no-load  current  out  of 


138  THE  ELECTRIC  CIRCUIT  [ART.  46 

consideration,  the  performance  of  the  motor  is  determined  by 
the  pure  magnetizing  current  /</  and  its  ratio  to  the  diameter  IL' 
of  the  semicircle.  This  ratio  is  called  the  circle  coefficient.  Let 
the  exciting  reactance,  or  the  reciprocal  of  60,  be  denoted  by  XQ. 
Then  /„'  =  EI/XO,  and,  by  definition,  the  circle  coefficient 

„  =  I0'/IL'  =  x/xo', (244) 

in  other  words,  the  circle  coefficient  is  equal  to  the  ratio  of  the 
leakage  reactance  to  the  exciting  reactance.1 

Thus,  knowing  the  magnetizing  current  and  the  short-circuit 
current  (or  the  magnetizing  current  and  the  circle  coefficient), 
the  Hey  land  circle  can  be  drawn,  and  the  relation  between  the 
primary  current,  the  load  current,  the  power-factor,  and  the 
angle  6  graphically  determined.  By  drawing  certain  auxiliary 
lines,  the  input,  output,  slip,  torque,  and  efficiency  can  also  be 
read  off  directly  from  the  diagram,  for  any  assumed  primary 
current.2  In  other  words,  the  circle  diagram  permits  one  to 
determine  graphically  the  performance  characteristics,  and  offers 
an  alternative  method  to  the  analytical  procedure  explained  in 
Art.  43  above.  The  relative  advantages  of  the  analytical  and 
graphical  methods  depend  upon  the  problem  in  hand  and  the 
skill  and  temperament  of  the  user;  the  student  should  thoroughly 
familiarize  himself  with  both  methods  before  deciding  upon  the 
use  of  one  or  the  other. 

The  circle  coefficient  is  very  convenient  for  preliminary  designs 
and  performance  estimates.  Mr.  H.  M.  Hobart  has  made  quite 
a  study  of  the  numerical  values  of  this  coefficient  for  a  large  num- 
ber of  actually  built  motors,  and  has  compiled  his  results  in  the 

1  The  circle  coefficient  is  also  called  the  dispersion  factor  (Streuungs- 
koeffizient),  and  is  usually  denoted   by  a.    Those  familiar  with  magnetic 
phenomena  will  notice  that  the  circle  coefficient  is  equal  to  the  ratio  of  the 
permeance  of  the  main  magnetic  path  in  the  motor  to  that  of  the  leakage 
paths.     This  is  because  the  reactances  are  proportional  to  the  corresponding 
inductances,  and  an  inductance  is  equal  to  the  permeance  of  the  path  times 
the  square  of  the  number  of  turns  linked  with  it.     A  motor  is  evidently  im- 
proved by  reducing  the  permeance  of  its  leakage  paths  and  increasing  that 
of  the  useful  path.    This  means  that  the  motor  is  better  the  lower  its  circle 
coefficient  a. 

2  For  complete  and  explicit  instructions  in  regard  to  the  construction  and 
use  of  the  circle  diagram,  see  the  author's  Experimental  Electrical  Engineer- 
ing, Vol.  2,  Chap.  29. 


CHAP.  XIII]  THE  INDUCTION   MOTOR  139 

form  of  charts,  from  which  the  value  of  the  coefficient  may  be 
taken  for  a  motor  of  given  or  assumed  dimensions.1 

Prob.  1.  Show  that  the  circle  coefficient  of  the  motor  specified  in 
problem  1,  Art.  43,  is  equal  to  0.0526. 

Prob.  2.  Check  a  few  points  on  the  curves  obtained  in  problem  1, 
Art.  43,  by  constructing  the  circle  diagram  of  the  motor. 

47.  The  Analytical  Determination  of  Performance.  —  Exact 
Solution.2  The  predetermination  of  the  performance  charac- 
teristics of  an  induction  motor,  explained  in  Arts.  43  and  46, 
is  based  upon  the  approximately  equivalent  diagram  shown  in 
Fig.  41.  The  exact  performance  characteristics  are  obtained  by 
expressing  analytically  the  electrical  relations  according  to  the 
correct  equivalent  diagram  shown  in  Fig.  42.  To  be  absolutely 
correct,  both  g0  and  b0  must  be  varied  somewhat  with  the  load, 
because  (a)  the  friction  and  windage  depend  upon  the  speed, 
(b)  the  iron  loss  is  not  exactly  proportional  to  the  square  of  the 
flux,  and  (c)  the  magnetizing  current  is  not  proportional  to  the 
voltage.  Moreover,  the  friction  loss  ought  to  be  separated  from 
the  iron  loss,  and  subtracted  from  the  output,  instead  of  being 
added  to  the  input.  All  these  corrections  make  the  calculations 
much  more  involved,  and,  while  it  is  well  to  know  about  them, 
they  are  hardly  ever  justified  in  practice. 

In  large  and  medium-sized  motors  the  losses  and  the  internal 
voltage  drop  are  comparatively  small,  so  that  the  performance 
calculated  according  to  the  exact  diagram  differs  but  little  from 
that  obtained  with  much  less  time  and  effort,  by  using  the 
approximate  diagram.  It  is  only  in  small  motors,  or  where  ex- 
treme accuracy  is  required  for  some  special  reason,  that  the  pro- 
cedure given  below  is  justified.  In  very  small  motors,  say  below 
one  kilowatt,  the  difference  between  the  approximate  and  the 
correct  performance  is  quite  appreciable,  because  of  high  losses 
and  a  large  voltage  drop. 

1  H.  M.  Hobart,  Electric  Molars  (1910),  Chapter  21.     It  may  be  of  inter- 
est to  note  that  the  correct  equivalent  diagram  (Fig.  42)  also  leads  to  a 
circle  diagram,  known  as  the  Ossanna  circle.     Numerous  articles  on  this 
exact  diagram  will  be  found  in  the  various  volumes  of  the  Elektrotechnische 
Zeitschrift  and  Elektrotechnik  und  Maschinenbau. 

2  This  article  may  be  omitted  if  desired,  because  the  approximate  solu- 
tion given  in  Arts.  43  and  46  is  sufficient  in  a  great  majority  of  practical  cases. 
However,  the  method  used  in  this  article  is  of  interest  to  the  student  as 
another  and  somewhat  different  application  of  complex  quantities. 


140  THE  ELECTRIC  CIRCUIT  [ART.  47 

It  is  much  more  convenient  to  plot  complete  performance 
curves  than  to  calculate  the  performance  data  for  a  specified  out- 
put. A  certain  external  resistance  R'  is  assumed,  such  as  would 
give  a  reasonable  value  of  slip  according  to  eq.  (214a),  and  the 
performance  characteristics  are  calculated  for  this  value  of  R'. 
Then  another  value  of  R'  is  assumed,  and  the  calculations  are 
repeated,  and  so  on.  For  an  assumed  value  of  R'  the  total 
admittance  between  the  primary  terminals  is  calculated,  using 
the  general  method  given  in  Art.  28,  that  is,  adding  impedances  in 
series  and  admittances  in  parallel.  Knowing  the  total  admittance, 
the  primary  current  becomes  known;  then  IL  and  EL  are  calcu- 
lated, and  finally  the  rest  of  the  data  are  obtained  as  in  Art.  43. 
The  details  of  the  calculations  are  as  follows : 

(1)  The  impedance  of  the  load  plus  that  of  the  secondary 
winding  =  (R'  +  r2')  +  jx-l . 

(2)  Using  eqs.  (121)  and  (122),  Art.  27,  find  the  corresponding 
admittance  02  —  jbz. 

(3)  The  total  admittance  between  the  points  M  and  N  is 
(go  4-  gz)  —j(b0  +  62). 

(4)  Using  eqs.  (123)  and  (124),  Art.  27,  find  the  corresponding 
impedance  rMN  +  JXMN> 

(5)  The  total  impedance  between  the  primary  terminals  "is 
Zeg  =  (TMN  +  n)  +  j  (XMN  +  #1). 

(6)  The  corresponding  admittance  Yeq  =  geq  -  jbeq  is  calcu- 
lated from  eqs.  (121)  and  (122). 

(7)  The  primary  current  is 

/i=#iFeg (245) 

(8)  The  voltage  across  MN 

#1  =  tfi  -  /xZi  =  ^(1- ZiF.,).    .     .     .     (246) 

(9)  The  load  current  is 

IL=  /!-  70=  /j-tfaFo, 

or,  substituting  the  values  of  /i  and  EH  from  eqs.  (245)  and  (246), 
IL  =  Ei  [Yeq  (1  +  FoZi)  -  Fo].       .     .     .     (247) 

(10)  The  load  voltage 

EL  =  lift' (248) 


CHAP.  XIII]  THE  INDUCTION  MOTOR  141 

The  rest  of  the  quantities  are  calculated  in  the  same  manner 
as  in  Arts.  43  and  44. 

In  numerical  work,  it  is  convenient  to  take  E\  along  the  ref- 
erence axis.  Having  determined  the  value  of  Yeq,  computations 
are  begun  with  the  composite  admittance  in  the  brackets  in 
eq.  (247).  Either  the  orthogonal  expressions  of  the  form  r  +  jx 
or  the  polar  expressions  of  the  form  z  (cos  0  +  j  sin  </>)  may  be 
used,  according  to  one's  preference  or  familiarity  with  one  or  the 
other  form.  The  student  ought  to  be  familiar  with  both  forms. 
The  trigonometric  form  is  convenient  for  multiplication  and 
division,  while  the  Cartesian  form  is  preferable  in  addition  and 
subtraction.  It  may  be  advisable  to  use  both  forms  in  the  same 
problem. 

The  calculator  should  avoid  long  algebraic  expressions,  jper- 
forming  numerical  operations  step  by  step.  Much  time  is  saved 
by  arranging  the  consecutive  steps  in  a  table,  so  as  to  repeat  the 
same  operations  mechanically  for  different  values  of  R'.  An 
irregularity  of  the  values  in  a  column  is  a  sure  indication  of  a 
numerical  error. 

Much  time  is  also  saved  by  intelligently  discriminating  be- 
tween the  principal  terms  and  small  correction  factors  in  an  ex- 
pression. For  instance,  in  eq.  (247)  Yeq  is  large  as  compared  to 
Yo  and  to  YeqYoZi.  It  would  be  a  waste  of  time  to  figure  out 
the  latter  expression  accurately,  when,  in  all  probability,  the 
principal  term  will  be  affected  only  by  its  first  significant  figure. 
On  the  other  hand,  the  principal  term,  Yeg,  must  be  calculated 
to  a  degree  of  accuracy  at  least  equal  to  that  desired  in  the  result, 
if  not  to  a  higher  degree.  Considerable  skill,  experience,  and  judg- 
ment are  necessary  to  determine  the  proper  accuracy  of  computa- 
tions in  engineering  problems.  This  is  an  art  which  grows  by 
intelligent  exercise,  and  it  is  never  too  early  to  begin  practicing 
it.  The  rewards  are  time  and  mental  energy  saved  for  better 
things,  while  obtaining  an  accuracy  which  is  commensurate  with 
the  desired  result.1 

1  For  a  complete  set  of  final  formulae  for  induction  motor  characteristics, 
see  Arnold's  Wechselstromtechnik,  Vol.  5,  part  1  (1909),  pp.  65-78.  A  very 
slight  inaccuracy  is  introduced  there  in  the  beginning,  by  neglecting  the 
imaginary  part  in  a  complex  quantity.  See  also  Dr.  Steinmetz's  Alternating- 
current  Phenomena,  under  "  Induction  Motor." 


142  THE  ELECTRIC  CIRCUIT  [ART.  47 

Prob.  1.  Make  out  a  table  showing  in  detail  the  order  of  computa- 
tions for  a  complete  set  of  performance  characteristics  of  an  induction 
motor,  according  to  the  method  developed  above. 

Prob.  2.  Mark  on  the  curve  sheet  obtained  in  problem  1,  Art.  43,  a 
few  points  determined  according  to  the  exact  equivalent  diagram,  in 
order  to  see  the  inaccuracy  resulting  from  the  use  of  the  approximate 
method. 


CHAPTER  XIV 
THE  DIELECTRIC  CIRCUIT 

48.  The  Electrostatic  Field.1  In  the  following  discussion,  it 
is  assumed  that  the  student  knows  the  fundamental  phenomena 
of  electrostatics  from  his  study  of  physics.  The  purpose  of  the 
treatment  given  here  is  to  deduce  the  principal  numerical  relations 
which  are  of  importance  in  electrical  engineering.  The  electro- 
static field  is  considered  in  this  book  from  Faraday's  point  of 
view,  viz.,  as  consisting  of  displacements  of  electricity,  and  stresses 


Battery 


Condenser 


FIG.  46.    A  plate  condenser  completing  a  direct- current  circuit. 

in  the  dielectric.     This  is  different  from  the  older  theory  of  the 
action  of  electric  charges  at  a  distance. 

Let  a  source  E  of  continuous  electromotive  force  (Fig.  46)  be 
connected  to  two  parallel  metallic  plates  A  and  B,  the  combina- 
tion of  which  is  commonly  known  as  a  condenser.  Let  the  plates 

1  See  the  footnote  at  the  beginning  of  Chapter  3. 
143 


144  THE  ELECTRIC  CIRCUIT  [ART.  48 

be  separated  from  each  other  by  air,  or  by  some  other  non-conduct- 
ing material.  When  the  key  K  is  pressed  upwards,  a  certain 
quantity  of  electricity,  Q,  flows  from  the  battery  to  the  plate  A, 
and  the  same  quantity  flows  from  the  plate  B  back  to  the  battery. 
This  quantity  can  be  measured  by  the  ballistic  galvanometer 
shown  in  the  circuit.  Within  a  very  short  time  the  difference  of 
potential  between  the  plates  becomes  equal  and  opposite  to  that 
of  the  battery  and  the  flow  of  current  stops. 

Since  electricity  behaves  like  an  incompressible  fluid,  the  same 
quantity,  Q,  is  displaced  through  the  whole  circuit,  including  the 
layer  of  insulation  or  dielectric  between  the  condenser  plates. 
This  displacement  is  accompanied  by  a  stress  in  the  dielectric, 
similar  in  some  respects  to  a  mechanical  stress  in  an  elastic  body. 
The  directions  of  the  electric  stress  and  of  the  lines  of  displacement 
of  electricity  through  the  air  are  shown  in  the  figure  by  dotted 
lines.  These  stresses  produce  a  counter-electromotive  force, 
which  finally  balances  that  of  the  battery.  When  the  key  is 
opened,  the  condenser  remains  charged,  since  the  stress  and  the 
displacement  can  be  relieved  only  in  a  closed  circuit.  To  dis- 
charge the  condenser,  its  plates  must  be  connected  by  a  conductor; 
this  is  done  by  pressing  the  key  down.  The  deflection  of  the 
ballistic  galvanometer  during  the  discharge  is  equal  and  opposite 
to  that  during  the  charge,  and  the  electric  energy  stored  in  the 
condenser  is  dissipated  by  the  current  in  the  form  of  heat. 

The  difference  between  a  dielectric  and  a  conductor  is  that  the 
resistance  of  the  former  to  the  passage  of  electricity  is  of  an  elastic 
nature;  that  is,  the  stress  can  be  relieved  and  the  stored  energy 
returned  to  the  circuit.  On  the  contrary,  the  resistance  to  the 
flow  of  electricity  in  a  conductor  is  of  the  nature  of  friction. 
The  energy  is  converted  into  Joulean  heat  and  cannot  be  restored. 

The  modern  electronic  theory  of  electricity  is  not  sufficiently 
advanced  at  this  writing  to  give  a  clear  account  of  the  true  nature 
of  these  displacements  and  stresses  in  a  dielectric.  It  is  therefore 
preferable  for  our  purposes  not  to  specify  the  mechanism  by  which 
these  stresses  and  displacements  are  produced.  We  shall  simply 
assume,  as  a  matter  of  fact,  the  structure  of  dielectrics  to  be  such 
that  an  e.m.f.  across  a  layer  of  such  material  produces  a  displace- 
ment of  a  certain  quantity  of  electricity,  which  is  proportional  to 
the  e.m.f.  When  the  e.m.f.  is  removed  and  a  closed  circuit  is 
provided,  the  stresses  within  the  dielectric  are  relieved,  and  the 


CHAP.  XIV]  THE  DIELECTRIC  CIRCUIT  145 

displacement  disappears.     The  analogy  to  an  elastic  body  sub- 
jected to  external  mechanical  forces  naturally  suggests  itself. 

Experiment  shows  that,  with  given  metallic  plates  (Fig.  46) 
and  the  same  applied  e.m.f.,  the  value  of  the  electric  displacement 
depends  upon  the  nature  of  the  dielectric.  With  solid  and  liquid 
insulating  materials,  such  as  glass,  oil,  mica,  etc.,  the  same  e.m.f. 
produces  larger  displacements  of  electricity  than  with  air  as  the. 
dielectric.  These  materials  are  therefore  said  to  possess  higher 
permittivity  than  the  air  (some  writers  use  the  word  inductivity) . 

When  an  alternating  voltage  is  applied  at  the  terminals  of  a 
condenser,  the  displacement  of  electricity  in  the  dielectric  varies 
continually  in  its  magnitude  and  periodically  reverses  its  direc- 
tion; consequently,  it  gives  rise  to  an  alternating  current  in  the 
conducting  part  of  the  circuit.  This  is  called  the  charging  or 
capacity  current.  This  current  leads  the  alternating  voltage  in 
phase  by  90  degrees,  as  may  be  seen  from  the  following  consider- 
ations: When  the  voltage  has  reached  its  instantaneous  maximum 
the  charging  current  is  zero,  because  at  the  crest  of  the  wave  the 
voltage  and  the  displacement  remain  practically  constant  for  a 
short  period  of  time.  As  soon  as  the  voltage  begins  to  decrease, 
the  current  begins  to  flow  in  the  direction  opposite  to  that  of  the 
applied  voltage,  because  the  elastic  reaction  of  the  dielectric  is 
now  larger  than  the  applied  electromotive  force.  At  any  instant, 
the  current,  or  the  rate  of  flow  of  electricity,  is  proportional  to 
the  rate  of  change  of  the  applied  voltage.  But  if  the  applied 
voltage  varies  according  to  the  sine  law,  the  rate  of  variation  is 
also  represented  by  a  sine  function  differing  in  phase  by  90  de- 
grees from  the  original  function,  because  d  (sin  x)/dx  =  cos  x  = 
sin  (90°+ x);  see  also  Art.  66  below.  That  there  must  be  a  dis- 
placement of  90  degrees  between  the  voltage  and  the  current 
follows  also  directly  from  the  assumed  elastic  structure  of  the 
dielectric.  The  energy  is  supposed  to  be  periodically  stored  in 
the  dielectric  and  given  up  again  without  any  loss;  hence,  the 
average  power  must  be  zero,  and  the  current  must  be  reactive. 

49.  A  Hydraulic  Analogue  to  the  Dielectric  Circuit.  The 
hydraulic  analogue  shown  in  Fig.  47  may  assist  the  student  in  the 
understanding  of  the  electrostatic  circuit.  A  is  a  pump  which 
corresponds  to  the  source  of  electromotive  force  in  Fig.  46.  The 
pipes  B  and  C  represent  the  leads  to  the  condenser,  or  the  metallic 
parts  of  the  circuit.  The  cylinder  D  corresponds  to  the  condenser, 


146 


THE  ELECTRIC  CIRCUIT 


[ART.  49 


and  the  elastic  partition  K  is  analogous  to  the  dielectric.  Let 
the  pipes  and  the  cylinders  be  filled  with  water,  and  let  the  piston 
in  A  be  in  its  middle  position,  the  partition  K  not  being  stressed. 
Let  the  stopcock  M  be  open,  and  the  stopcock  N  closed.  When  a 


Stop  Cocks 


FIG.  47.    A  hydraulic  analogue  of  a  dielectric  circuit. 

pull  to  the  right  is  exerted  upon  the  piston  rod  and  it  is  forced 
to  move,  the  water  in  the  system  is  displaced,  and  the  elastic 
partition  K  is  strained,  as  shown  in  the  figure.  With  a  given 
pull,  or  a  given  electromotive  force,  the  movement  stops  when 
the  pull  is  balanced  by  the  elastic  reaction  of  the  partition. 
The  charge,  or  the  total  displacement,  is  represented  by  the 
amount  of  water  shifted;  it  can  be  measured  by  the  water-meter 
W,  which  thus  takes  the  place  of  the  ballistic  galvanometer. 

If  the  pipes  are  frictionless,  and  the  inertia  of  the  piston  and 
water  is  assumed  negligible,  the  analogy  can  be  followed  still 
further;  namely,  the  phase  difference  in  time  between  the  pull 
and  the  velocity  of  the  water  is  equal  to  90  degrees,  the  velocity 
leading  the  pull.  Assuming  the  motion  of  the  piston  to  be  har- 
monic, the  velocity  of  the  flow  of  water  is  at  its  maximum  when 
the  piston  is  at  the  center  of  its  stroke.  The  required  pull  is  equal 
to  zero  at  this  moment,  because  the  elastic  partition  is  in  its 
middle,  or  unstrained  position.  At  the  end  of  the  stroke  the 
velocity  is  zero,  but  the  pull  is  at  its  maximum,  because  the 
partition  is  strained  to  its  extreme  position,  and  exerts  its  maxi- 
mum elastic  reaction.  Thus  the  pull  lags  behind  the  velocity. 

Substituting   another   partition,    made   of   a   more   yielding 


CHAP.  XIV]  THE  DIELECTRIC  CIRCUIT  147 

material  (material  possessing  higher  permittivity),  a  larger  dis- 
placement is  produced  with  the  same  pull;  this  corresponds  to 
the  case  in  which  some  solid  or  liquid  dielectric  is  substituted  for 
the  air. 

Closing  the  stopcock  M  corresponds  to  breaking  the  electric 
circuit  of  the  condenser.  It  will  be  seen  from  analogy  that  the 
condenser  remains  charged.  To  discharge  the  condenser,  the 
stopcock  N  must  be  opened;  this  equalizes  the  pressure  on  both 
sides  of  the  elastic  partition.  Since,  in  reality,  water  possesses 
some  inertia,  the  partition  does  not  stop  in  its  middle  position 
during  the  discharge,  but  the  momentum  of  the  water  carries  it 
beyond  the  center.  The  electromagnetic  inertia  of  the  electric 
current  produces  a  similar  effect,  and  we  thus  have  a  simple 
explanation  of  the  oscillatory  character  of  the  electric  discharge. 
During  this  discharge,  the  energy  is  alternately  transformed  into 
the  potential  energy  of  dielectric  stress,  and  into  kinetic  energy  of 
the  magnetic  field.  The  oscillations  of  the  partition  are  gradually 
damped  out  by  the  frictional  resistance  of  the  pipes.  In  the 
electric  circuit,  oscillations  are  damped  by  the  ohmic  resistance 
of  the  conducting  parts  of  the  circuit. 

The  student  can  follow  this  analogy  still  further,  and  exp^iin 
free  electrical  vibrations,  current  and  voltage  resonance,  also  the 
effect  of  a  resistance  in  series  and  in  parallel  with  a  condenser,  etc. 

50.  The  Permittance  and  Elastance  of  Dielectric  Paths. 
Let  Q  (Fig.  46)  be  the  total  displacement  of  electricity  in  the 
dielectric,  measured  in  ampere-seconds  or  coulombs,  and  let  E 
be  the  voltage  impressed  across  the  condenser  or  "  permittor." 
Experiment  shows  that  up  to  a  certain  limit  Q  is  proportional  to 
E;  this  is  similar  to  the  behavior  of  an  elastic  body,  in  which  the 
strains  are  proportional  to  the  applied  forces  until  the  limit  of 
elasticity  has  been  reached.  Thus,  we  may  write 

Q  =  CE, (249) 

where  the  coefficient  of  proportionality,  C,  is  called  the  permittance 
of  the  condenser.  The  older  name  for  C  is  electrostatic  capacity. 
When  E  is  in  volts  and  Q  in  coulombs,  permittance  is  measured 
in  units  called  farads.  A  condenser  has  a  permittance  of  one 
farad  when  a  displacement  of  one  coulomb  is  produced  for  each 
volt  applied  at  its  terminals.  The  farad  being  too  large  a  unit 
for  practical  use,  permittances  are  usually  measured  in  micro- 


148  THE  ELECTRIC  CIRCUIT  [ART.  50 

farads,  one  microfarad  being  equal  to  one  millionth  part  of  a 
farad. 

The  larger  the  permittance  of  a  condenser,  the  larger  is  the 
displacement  of  electricity  with  the  same  voltage;  hence  C  is  a 
measure  of  the  ease  with  which  an  electric  displacement  can  be 
produced  in  a  given  condenser.  In  this  respect  the  concept  of 
permittance  is  analogous  to  those  of  electric  conductance  and 
magnetic  permeance. 

In  some  cases  it  is  convenient  to  speak,  not  of  the  degree  of 
ease,  but  of  the  difficulty  with  which  an  electric  displacement 
can  be  produced  in  a  given  condenser.  For  this  purpose,  a 
coefficient  of  proportionality,  the  reciprocal  of  C,  has  to  be  used; 
and  eq.  (249)  becomes 

E  =  SQ, (250) 

where 

S  =  C-1       (251) 

is  called  the  elastance  of  the  condenser.  Elastance  is  thus  analo- 
gous to  electric  resistance  and  to  magnetic  reluctance.  When 
permittance  is  measured  in  farads,  the  unit  of  elastance  is  the 
reciprocal  of  the  farad,  and  may  therefore  be  properly  called  the 
daraf.  This  is  a  name  derived  by  spelling  the  word  farad  back- 
wards, that  is,  in  the  same  way  in  which  mho  is  derived  from  ohm.1 
A  condenser  has  an  elastance  of  one  daraf  when  one  volt  of  pres- 
sure is  required  for  each  coulomb  of  displacement  within  it.  The 
farad  being  too  large  a  unit  for  practical  use,  the  daraf  is  con- 
sequently too  small  a  unit.  Therefore,  in  practice,  elastances 
should  be  measured  in  megadarafs,  one  megadaraf  (=  106  darafs) 
being  the  reciprocal  of  one  microfarad.2 

When  two  or  more  permittances  are  connected  electrically 
in  parallel,  the  resultant  permittance  is  larger  than  that  of  any 
of  the  component  condensers,  because  a  larger  path  is  offered  to 

1  It  may  be  of  interest  to  mention  in  this  connection  a  similar  derivation 
of  the  name  for  a  unit  of  magnetic  reluctance.    The  henry  being  the  natural 
unit  of  magnetic  permeance  (or  inductance)  in  the  ampere-ohm  system,  the 
author  has  proposed  calling  the  corresponding  unit  of  reluctance  the  yrneh,  a 
word  derived  by  spelling  the  word  henry  backwards.     See  his  Magnetic  Circuit, 
Art.  5. 

2  For  a  complete  rational  nomenclature  of  electric  and  magnetic  quanti- 
ties, see  the  table  on  page  xii  at  the  beginning  of  the  book,  and  also  the  one  in 
the  Appendix. 


CHAP.  XIV]  THE  DIELECTRIC  CIRCUIT  149 

the  displacement.  The  relation  is  similar  to  that  of  conductances 
or  permeances  in  parallel.  Let  d,  C2,  etc.,  represent  permit- 
tances connected  in  parallel  across  a  source  of  constant  voltage 
E,  and  let  Qi,  Q2,  etc.,  be  the  corresponding  electric  displacements 
through  these  condensers  (or  permittors).  Then,  according  to 
the  definition  of  permittance,  we  have 

Qi  =  C,E 


(252) 


The  equivalent  permittance,  Ceq,  must  be  such  as  to  allow  of  a 
displacement  equal  to  the  sum  of  the  partial  displacements,  with 
the  same  voltage;  hence, 

SQ  =  CeqE (253) 

Adding  eqs.  (252)  together  gives 

Qi  +  Q2  +  etc.  =  E  (Ci  +  Cz  +  etc.), 
or,  by  comparison  with  eq.  (253), 

C.,  =  2C (254) 

In  other  words,  when  permittances  are  connected  in  parallel, 
the  equivalent  permittance  is  equal  to  their  sum. 

When  condensers  (or  elastors)  are  connected  in  series,  it  is 
more  convenient  to  use  their  elastances.  Since  electricity  behaves 
like  an  incompressible  fluid,  the  displacement  through  several 
elastances  in  series  is  the  same  in  all  of  them.  Let  this  displace- 
ment be  denoted  by  Q,  and  let  the  voltages  across  the  terminals 
of  the  individual  elastors  be  Eif  Ez,  etc.  Then, 


(255) 


where  Si,  Sz,  etc.,  are  the  elastances  of  the  separate  condensers. 
The  equivalent  elastance  must  allow  of  the  same  displacement  Q 
with  the  same  total  voltage,  or 

2E  =  SegQ.    .......     (256) 

Adding  eqs.  (255)  together  gives 

Ei  +  Ez  +  etc.  =  Q  (Si  +  S2  +  etc.), 


150  THE  ELECTRIC  CIRCUIT  [Aux.  51 

or,  by  comparison  with  eq.  (256), 

Seq  =  2S.      . (257) 

In  other  words,  when  elastances  are  connected  in  series,  the 
equivalent  elastance  is  equal  to  their  sum.  The  analogy  to  the 
addition  of  conductances  in  parallel  and  resistances  in  series  is 
self-evident  (see  Art.  3). 

Prob.  1.  A  condenser,  which  has  a  permittance  of  10  microfarads,  is 
connected  to  a  direct-current  magneto,  the  speed  of  which  is  increased  at 
a  uniform  rate,  so  that  the  voltage  rises  at  a  rate  of  1.7  volts  per  second. 
Calculate  the  charging  current. 

Ans.  17  microamperes.  Note:  This  is  the  principle  of  an  appara- 
tus used  for  measuring  the  acceleration  of  railway  trains. 

Prob.  2.  An  elastance  of  10  kilodarafs  is  connected  across  a  220- 
volt,  50-cycle  line.  Show  that  the  effective  value  of  the  charging  current 
is  6.91  amp.  Solution:  The  maximum  displacement  in  the  dielectric  is 
220  V2/(10  X  103)  =  22  V2  X  10~3  coulombs.  This  displacement  is 
reduced  to  zero  within  aU  of  a  second;  hence,  the  average  charging 
current  is  4.4  \/2  amp.  The  effective  value,  assuming  a  sine-wave  of 
current,  is  4.4  v72  X  (-**•/ V2)  -  6.91  amp. 

Prob.  3.  Show  that  with  two  condensers  in  parallel  the  ratio  of  the 
displacements  equals  that  of  the  permittances  or  is  inversely  as  the  ratio 
of  the  elastances.  What  is  the  analogous  relation  for  conductances  and 
resistances? 

Prob.  4.  When  two  condensers  are  in  series,  show  that  the  ratio  of 
the  voltage  drops  across  them  equals  that  of  the  elastances,  or  is  inversely 
as  the  ratio  of  the  permittances.  What  is  the  analogous  relation  for 
resistances  and  conductances? 

Prob.  6.  A  sectionalized  condenser,  such  as  is  used  for  calibration 
and  exact  measurements,  is  built  up  of  the  following  permittances: 
0.5,  0.2,  0.2,  0.05,  and  0.05  microfarads.  What  is  the  extreme  range 
of  permittances  and  elastances  possible  by  combining  these  sections  in 
series  and  in  parallel? 

Ans.  From  1  to  0.0192  mf.,  or  from  1  to  52  mgd. 
Prob.  6.  Referring  to  the  preceding  problem,  the  sections  of  the  con- 
denser are  connected  as  follows:  0.2,  0.05,  and  0.05  mf.  are  in  scries,  and 
the  combination  is  shunted  by  0.2  mf.  Then  the  whole  is  put  in  series 
with  0.5  mf.  Show  that  the  resultant  permittance  is  equal  to  0.154 
microfarads. 

51.  Permittivity  and  Elastivity  of  Dielectrics.  Experiment 
shows  that  the  permittance  of  a  sample  of  any  dielectric  varies 
with  its  dimensions  in  the  same  way  that  the  conductance  of  a 
metal  or  the  permeance  of  a  magnetic  path  in  a  non-ferrous  medium 
does;  namely,  the  permittance  is  proportional  to  the  cross-sec- 
tion of  the  layer  and  inversely  proportional  to  its  length  in  the 


CHAP.  XIV]  THE  DIELECTRIC  CIRCUIT  151 

direction  of  the  lines  of  force.  By  increasing  the  cross-section  of 
the  path  perpendicular  to  the  lines  of  force  (Fig.  46),  the  displace- 
ment is  increased  in  the  same  proportion.  On  the  other  hand, 
the  displacement  is  found  to  be  inversely  proportional  to  the 
thickness  of  the  dielectric,  since  the  distance  through  which  the 
voltage  must  act  is  greater  if  the  thickness  is  increased.  These 
relations  follow  directly  from  the  laws  deduced  in  the  preceding 
article  for  the  addition  of  permittances  in  parallel  and  elastances 
in  series.  Thus,  by  analogy  with  eq.  (21),  Art.  5,  we  put 

C  =  KA/l, (258) 

where  K  is  called  the  permittivity  of  the  dielectric.  It  is  analogous 
to  the  conductivity  of  a  conducting  material,  or  to  the  perme- 
ability of  a  magnetic  medium.  Permittivity  may  be  defined  as 
the  permittance  of  a  cubic  unit  of  dielectric,  when  the  lines  of 
displacement  are  straight  lines  perpendicular  to  one  of  its  faces. 
For  air  the  permittivity  is 

K«  =  0.08842  X  10"6  microfarads  per  cm.  cube.  .  (259) 
For  other  dielectrics,  liquid  and  solid,  the  permittivity  is  higher  than 
that  of  air;  that  is  to  say,  they  are  more  yielding  to  an  electro- 
motive force.  It  is  convenient  to  express  their  permittivities  in 
terms  of  that  of  the  air;  for  instance,  we  may  say  that  the  per- 
mittivity of  a  certain  transformer  oil  is  2.1  times  that  of  the  air. 
The  relative  permittivities  of  some  important  insulating  ma- 
terials are  tabulated  in  Art.  56  below,  merely  to  indicate  their 
order  of  magnitude.  For  accurate  values,  the  reader  is  referred 
to  various  published  physical  tables  and  engineering  handbooks. 
The  older  name  for  relative  permittivity  is  specific  inductive 
capacity  (or  dielectric  constant).  It  is  more  convenient  in  prac- 
tice to  use  relative  than  absolute  permittivities,  because  the 
necessity  of  tabulating  small  quantities  like  *„  in  eq.  (259)  is 
avoided.  Besides,  the  data  are  more  readily  comparable  with 
one  another,  and  with  the  permittivity  of  air,  which  is  a  standard 
dielectric.  This  procedure  is  analogous  to  tabulating  the  con- 
ductivities of  various  metals  in  terms  of  that  of  pure  copper, 
taken  as  100  per  cent.  The  absolute  permittivity  of  a  material 
is  obtained  by  multiplying  the  absolute  permittivity  of  air  by  the 
relative  permittivity  of  the  dielectric  in  question.  Equation  (258) 
thus  becomes 

C  =  KKa  A/I, (260) 

where  K  stands  for  the  relative  permittivity. 


152  THE  ELECTRIC  CIRCUIT  [ART.  51 

The  clastance  of  a  prismatic  piece  of  dielectric,  with  the  lines 
of  displacement  parallel  to  one  set  of  its  edges,  is  expressed  by 
analogy  with  eq.  (20),  Art.  5,  as 

S  =  ffl/A, (261) 

where 

a  =  K~l (262) 

is  called  the  elastivity  of  the  dielectric.  Elastivity  is  analogous  to 
the  resistivity  of  a  conducting  material  or  to  the  reluctivity  of  a 
magnetic  medium,  and  may  be  expressed  for  practical  purposes  in 
megadarafs  per  centimeter  cube.  For  air,  the  absolute  elastivity 
is,  according  to  eq.  (259), 

ffa  =  Ka-i  =  11.3  x  106  megadarafs  per  cm.  cube.  .  (263) 
The  concept  of  relative  elastivity  could  be  introduced  if  necessary, 
in  which  case  its  values  would  be  equal  to  the  reciprocals  of  the 
relative  permittivities  tabulated  in  Art.  56.  However,  it  is  suf- 
ficient to  use  the  relative  permittivity,  even  when  dealing  with 
elastances,  so  that  eq.  (261)  becomes 

S  =  (ffa/K)l/A (264) 

The  nomenclature  used  above  is  due  to  Mr.  Heaviside;1  it  is 
consistent  and  uniform  with  the  nomenclature  used  in  the  electro- 
conducting  and  magnetic  circuits,  and  is  suggestive  as  to  the  nature 
of  the  phenomena.  The  electrostatic  nomenclature  now  in  general 
use  comprises  but  three  terms;  namely,  condenser,  capacity,  and 
specific  inductive  capacity.  It  is  hoped  that  the  more  rational  and 
complete  nomenclature  used  here  will  help  to  a  clearer  understand- 
ing of  the  dielectric  circuit,  and  will  simplify  engineering  calcula- 
tions relating  thereto.2 

Note:  The  author  considers  the  above-given  value  of  «0,  eq.  (259), 
to  be  an  experimental  coefficient,  in  the  same  sense  in  which  other  prop- 
erties of  materials  are  characterized  by  experimental  coefficients.  For  an 
engineer,  the  volt  and  the  ampere  are  arbitrary  units  established  by  an 
international  agreement,  no  matter  what  their  relation  to  the  so-called 
absolute  units.  The  value  of  «„  can  be  calculated  theoretically,  assuming 
the  ratio  between  the  electrostatic  and  the  electromagnetic  units  to  be 
known.  In  the  absolute  electrostatic  system  of  units,  with  air  as  the 
dielectric,  a  plate  condenser  having  an  area  of  A  sq.  cm.  and  a  distance 
between  the  plates  equal  to  I  cm.,  has  a  capacity  equal  to  A/ (4*1).  The 

1  O.  Heaviside,  Electromagnetic  Theory  (1894),  Vol.  1,  p.  28. 

1  See  the  author's  paper  "  Sur  Quelques  Calculs  Pratiques  des  Champs 
Electrostatiques,"  in  the  Transactions  of  the  Congresso  Internazionak  delle 
Applicazioni  ElcUriche,  Turin,  1911. 


CHAP.  XIV]  THE  DIELECTRIC  CIRCUIT  153 

factor  4  w  enters  on  account  of  an  unfortunate  selection  of  the  expression 
for  Coulomb's  law,  which  should  have  been  gig2/4  Trr2,  instead  of  qiq^/r2. 
In  the  absolute  electromagnetic  units  the  same  capacity  is  equal  to 
(A/4  Ti-Z)  (3  X  lO10)"2,  where  3  X  1010  is  the  velocity  of  light  in  centimeters 
per  second.  To  obtain  the  result  in  microfarads,  the  foregoing  expression 
must  be  multiplied  by  1015.  On  the  other  hand,  the  same  capacity 
expressed  in  the  rational -units  defined  above  is  KaA/l.  Equating  the  two 
expressions  gives  Ka  =  lQ-5/(9  X  4  *•)  =  0.08842  X  lO"6  microfarads  per 
centimeter  cube. 

The  fact  that  Ka  can  be  expressed  through  the  velocity  of  light  does 
not  make  *«,  the  less  an  empirical  coefficient,  because  the  velocity  of  light 
itself  is  determined  experimentally.  As  a  matter  of  fact,  one  of  the  ways 
in  which  the  velocity  of  light  is  determined  consists  in  calculating  it 
indirectly  from  the  value  of  «a  obtained  from  measurements. 

Prob.  1.  Show  that  in  the  English  system  *a  =  0.2244  X  10~6  micro- 
farads per  inch  cube. 

Prob.  2.  A  condenser  (Fig.  46)  consists  of  two  metal  plates,  50  by 
70  cm.  each,  in  contact  with  a  glass  plate  3  mm.  thick  between  them. 
When  a  continuous  voltage  of  2400  is  applied  to  the  condenser,  the  ballis- 
tic galvanometer  shows  a  charge  of  17.1  microcoulombs.  What  is  the 
relative  permittivity  of  the  glass?  Ans.  6.9 

Prob.  3.  A  0.5-mf .  mica  condenser  is  to  be  made  out  of  sheets  of  mica 
12  by  25  cm.,  0.3  mm.  thick,  and  coated  on  one  side  with  a  very  thin 
film  of  silver.  How  many  sheets  are  required?  The  relative  permittivity 
of  the  mica  is  about  6. 

Ans.    About  96  sheets,  48  sheets  in  parallel  per  terminal. 

Prob.  4.  Let  the  dielectric  in  problem  2  consist,  instead  of  glass,  of 
three  layers  of  different  materials.  Let  the  thicknesses  of  these  layers 
be  1.2,  0.7,  and  1.1  mm.,  and  let  the  corresponding  values  of  relative  per- 
mittivities be  2,  3,  and  5.  What  is  the  capacity  of  the  condenser?  Hint: 
Calculate  the  equivalent  elastance  as  the  sum  of  three  elastances  in  series. 

Ans.    2.94  X  10~3  mf. 

52.  Dielectric  Flux  Density  and  Electrostatic  Stress  (Voltage 
Gradient).  Referring  again  to  the  uniform  electrostatic  field 
(Fig.  46),  consider  a  cube  of  the  dielectric,  one  square  centimeter 
in  cross-section,  and  one  centimeter  long  in  the  direction  of  the 
lines  of  force.  Let  a  quantity  of  electricity  Q  be  supplied  by  the 
battery,  as  shown  by  the  ballistic  galvanometer;  then  the  same 
quantity  of  electricity  must  be  displaced  in  the  dielectric.  Neg- 
lecting a  small  displacement  at  the  edges  and  at  the  outside  sur- 
faces of  the  plates,  the  whole  quantity  Q  is  uniformly  displaced 
between  the  plates.  Therefore,  if  the  area  of  each  plate  is  equal 
to  A  square  centimeters,  the  displacement  through  the  cube  under 
consideration  is  equal  to 

D  =  Q/A (265) 


154  THE  ELECTRIC  CIRCUIT  [ART.  52 

Since  Q  is  the  total  electrostatic  flux,  D  is  naturally  called  the 
dielectric  flux  density.  If  Q  is  measured  in  coulombs,  D  is  ex- 
pressed in  coulombs  per  square  centimeter.  In  practice,  Q  is 
measured  in  microcoulombs,  and  D  is  expressed  in  microcoulombs 
per  square  centimeter.  The  dielectric  flux  density  is  analogous 
to  current  density  U  (Art.  6)  and  to  magnetic  flux  density  B. 

When  an  electrostatic  field  is  non-uniform  (Fig.  48),  it  is  con- 
veniently subdivided  by  lines  of  force  and  equipotential  surfaces 
perpendicular  to  the  same.  The  procedure  is  similar  to  that  used 
in  Art.  8.  In  this  case,  the  total  flux  or  displacement  divided  by 
the  area  of  an  equipotential  surface  gives  only  the  average  flux 
density  through  the  surface.  The  actual  density  varies  from 
point  to  point,  and  it  is  therefore  proper  to  speak  of  the  dielectric 
flux  density  at  a  point.  Take  a  tube  of  infinitesimal  cross-section 
formed  by  lines  of  force,  and  let  dQ  be  the  displacement  of  electric- 
ity through  this  tube.  The  displacement  is  the  same  through 
any  normal  cross-section  of  the  tube,  because  electricity  behaves 
like  an  incompressible  fluid.  Let  dA  be  a  particular  cross-section 
of  the  tube;  then  the  flux  density  at  this  cross-section  is 

D  =  dQ/dA, (266) 

D  being  usually  expressed  in  coulombs  (or  microcoulombs)  per 
square  centimeter.  Since  the  cross-section  of  the  tube  is  infinites- 
imal, D  is  the  density  at  the  point  corresponding  to  the  position 
of  dA. 

If  the  flux  density  in  a  uniform  field  is  given,  the  total  displace- 
ment is 

Q  =  DA (267) 

In  a  non-uniform  field,  the  flux  density  must  be  given  as  a  function 
of  the  coordinates  of  the  field;  so  that 

Q  =    f    DdA,     .  (268) 

Jo 

the  integration  being  extended  over  the  whole  area  of  an  equi- 
potential surface,  or  over  the  part  of  this  area  through  which  the 
flux  is  to  be  calculated. 

The  electromotive  force  impressed  at  the  terminals  of  a 
condenser  is  balanced  in  the  whole  thickness  of  the  dielectric; 
that  is,  each  small  length  of  path  in  the  dielectric  produces  its 
own  counter-electromotive  force.  Therefore,  it  is  possible  to 
speak  of  the  voltage  drop  per  unit  length  of  the  path  in  the 


CHAP.  XIV] 


THE  DIELECTRIC  CIRCUIT 


155 


dielectric,  the  same  as  in  Art.  6.     This  voltage  gradient,  or  electric 
intensity,  in  a  uniform  field  is  expressed  by 

G  =  E/l, (269) 

and  is  measured,  as  in  the  conducting  circuit,  in  volts  per  centi- 
meter, kilovolts  per  millimeter,  or  in  other  suitable  units. 

In  a  non-uniform  field,  the  electric  intensity,  or  voltage  gradi- 
ent, varies  from  point  to  point.  Let  the  voltage  between  two 
infinitely  close  equipotential  surfaces  MN  and  M'N'  (Fig.  48) 


FIG.  48.    A  non-uniform  electrostatic  field,  represented  by  lines  of 
displacement  and  equipotential  surfaces. 

be  dE,  and  let  the  distance  mn  between  the  surfaces,  along  a 
certain  line  of  force  HH',  be  dl.  Then  the  voltage  gradient  along 
mn  is 

G  =  dE/dl (270) 

The  length  of  the  line  mn  being  infinitesimal,  G  is  the  intensity  at 
any  point  between  m  and  n. 

When  the  voltage  gradient  is  uniform,  we  have  for  the  total 
voltage  across  the  field 

E  =  Gl (271) 

In  a  non-uniform  field,  G  has  to  be  given  as  a  function  of  I,  so  that 


E  = 


(272) 


156  THE  ELECTRIC  CIRCUIT  [ART.  52 

the  integration  being  performed  between  any  two  points  on  the 
equipotential  surfaces  between  which  the  voltage  is  to  be  deter- 
mined. 

Imagine  a  uniform  field  existing  in  a  dielectric,  and  consider  a 
unit  cube  of  the  material.  The  total  displacement  through  such 
a  cube  is  equal  to  the  flux  density  D,  and  the  voltage  across  it  is 
equal  to  the  voltage  gradient  G.  The  permittance  and  the  elas- 
tance  of  the  cube  are  respectively  equal  to  the  permittivity  and 
the  elastivity  of  the  material.  Thus,  applying  to  the  cube 
eqs.  (249)  and  (250),  we  have 

D  =  KG, (273) 

and 

G  =  <rD (274) 

These  equations  are  analogous  to  eq.  (25)  in  Art.  6  of  this  book 
and  to  eqs.  (15)  and  (16),  Art.  8,  of  the  Magnetic  Circuit.  These 
relations  may  be  considered  as  fundamental  in  the  theory  of  the 
dielectric  circuit,  G  being  the  cause,  D  the  effect,  and  K  (or  a) 
the  coefficient  of  proportionality  which  characterizes  the  material. 
Similar  linear  relations  between  cause  and  effect  holS  in  the  con- 
duction of  heat,  and  in  the  theory  of  elasticity.  The  voltage 
gradient  is  sometimes  called  the  stress  in  the  dielectric,  eq.  (274)  being 
analogous  to  Hooke's  law  for  elastic  bodies.  The  elastivity  a  takes 
the  place  of  the  modulus  of  elasticity. 

If  the  field  or  the  dielectric  is  non-uniform,  eqs.  (273)  and  (274) 
still  hold  true  for  every  point,  D  being  the  dielectric  flux  density 
and  G  the  voltage  gradient  at  the  point  considered.  This  can  be 
proved  by  applying  eqs.  (249)  and  (250)  to  an  infinitesimal  par- 
allelopiped  instead  of  a  unit  cube. 

Equations  (268)  and  (272)  are  expressed  in  words  by  saying 
that  the  total  displacement  Q  is  the  surface  integral  of  the  dielectric 
flux  density  D,  and  the  voltage  E  is  a  line  integral  of  the  gradient 
or  stress  G.  These  statements  are  almost  self-evident  from  the 
definition  of  the  quantities  and  the  structure  of  dielectrics. 

Prob.  1.  What  arc  the  dielectric  flux  density  and  voltage  gradient  in 
problem  2,  Art.  51?  Ans.  4.885  X  lO"3  nrc./cm.2;  8  kv./cm. 

Prob.  2.  The  condenser  specified  in  problem  4,  Art.  51,  is  subjected 
to  a  difference  of  potential  of  10  kv.  What  are  the  voltage  gradients 
(stresses)  in  the  three  layers  of  dielectric? 

Ans.    4.75;  3.16;   1.9  kv./mm. 


CHAPTER  XV 
THE  DIELECTRIC  CIRCUIT  —  (Continued) 

53.  Energy  in  the  Electrostatic  Field.  When  a  dielectric 
is  being  charged,  a  current  flows  into  it  from  the  source  of  elec- 
tromotive force.  This  involves  the  expenditure  of  a  certain 
amount  of  energy,  because  the  counter-e.m.f.  due  to  the  dielectric 
stresses  has  to  be  overcome.  This  energy  is  not  converted  into 
heat,  and  lost,  as  in  the  case  of  metallic  conduction:  it  is  stored 
in  the  dielectric  in  potential  form,  and  can  be  returned  to  the  cir- 
cuit by  reducing  the  voltage  at  the  condenser  terminals.  With 
reference  to  the  analogy  shown  in  Fig.  47,  the  mechanical  energy 
expended  by  the  pump  in  straining  the  elastic  partition  is  stored 
in  the  partition,  in  the  form  of  potential  energy.  This  energy 
can  be  returned  to  the  piston  rod  by  allowing  it  to  be  moved  by 
the  elastic  forces  of  the  partition. 

In  some  cases  it  is  necessary  to  calculate  the  energy  stored  in 
an  electrostatic  field;  or  to  express  the  energy  stored  per  cubic 
centimeter  of  dielectric,  as  a  function  of  the  stress  G  and  flux 
density  D,  at  the  point  under  consideration. 

Consider  first  the  simple  case  of  a  uniform  field  (Fig.  46), 
and  neglect  the  small  amount  of  displacement  occurring  outside 
the  space  between  the  plates.  Let  the  dielectric  be  charged  by 
gradually  raising  the  voltage  between  its  limiting  surfaces  from 
zero  to  a  final  value  E;  and  let  e  and  i  be  the  instantaneous  values 
of  the  voltage  and  charging  current  at  a  moment  t  during  the 
process  of  charging.1  The  total  electrical  energy  delivered  to  the 
dielectric  in  charging  it  is 


f*T  f*T 

W  =    I     eidt=    I    e>dq,      . 
Jo  Jo 


(275) 


where  T  is  the  total  time  of  charging,  and  dq  =  i  dt  is  the  infini- 
tesimal charge  or  displacement  added  to  the  condenser  during 
the  interval  of  time  dt.  The  quantities  dq  and  e  can  be  expressed 

1  The  voltage  and  the  charging  current  rise  gradually,  even  though  the 
key  K  be  closed  suddenly.  This  is  on  account  of  an  ever-present  magnetic 
inductance  which  acts  as  a  kind  of  electromagnetic  inertia. 

157 


158  THE  ELECTRIC  CIRCUIT  [ART.  53 

through  the  instantaneous  flux  density  Dt  and  the  stress  Gt', 
namely,  from  eq.  (267),  dq  =  A  dDt,  and  from  eq.  (271)  e  =  Gtl. 
Performing  the  substitution,  and  taking  the  constant  quantities  A 
and  I  outside  of  the  sign  of  integration,  we  get 

=  Al  CTGtdDt  ......     (276) 

Jo 

In  order  to  integrate  this  expression,  Dt  must  be  expressed  through 
Gt,  or  vice  versa.  The  relation  between  the  two  is  given  by  eq. 
(273).  Eliminating  Dt,  we  obtain 

W 


W 


=  KAl  fTGt  dGt  =  \  KVG\       .     .     .     (277) 


where  V  =  A  I  is  the  volume  of  the  dielectric,  and  G  is  the  final 
value  of  the  stress,  at  the  time  T.  Hence,  the  energy  stored  per 
unit  volume  of  the  dielectric,  or  the  density  of  energy,  is 

W'  =  W/V  =  \  KG2  =  \  G*/a  .....     (278) 

Using  relations  (273)  and  (274),  the  preceding  formula  can  also 
be  written  in  the  following  forms: 

W'  =  i  GD  =  i  £2/K  =  \  aD*  .....     (279) 

The  analogy  to  the  corresponding  formula  in  Art.  69  of  the  Mag- 
netic Circuit  is  apparent  at  once. 

The  total  stored  energy  can  be  expressed  through  the  per- 
mittance or  elastance  of  the  dielectric.  We  have  from  eq.  (249) 
dq  =  C'de;  substituting  in  eq.  (275)  and  integrating,  we  get 

W  =  \  CE*  =  i  E2/S  ......     (280) 

Since  the  final  charge,  or  total  displacement  Q  equals  CE  or 
E/S,  the  energy  can  be  represented  also  in  the  following  forms: 
W  =  i  QE  =  \  Q*/C  =  \  Q*S  .....     (281) 

These  formulae  are  analogous  to  the  corresponding  expressions  in 
Art.  57  of  the  Magnetic  Circuit. 

Let  now  the  dielectric  and  the  field  be  of  an  irregular  form 
as  shown  in  Fig.  48.  The  stress  G  and  the  displacement  D  are 
different  at  different  points,  so  that  it  is  necessary  to  consider 
infinitesimal  layers  of  the  dielectric  between  consecutive  equi- 
potential  surfaces,  and  infinitesimal  threads  of  displacement  be- 
tween the  electrodes.  Consider  an  infinitesimal  volume  mnqp  of 
the  dielectric,  comprising  the  part  of  a  tube  of  displacement  EH' 
between  two  equipotential  surfaces  MN  and  M'N'.  The  sides 


CHAP.  XV]  THE  DIELECTRIC  CIRCUIT  159 

mp  and  nq  can  be  provided  with  infinitely  thin  metal  films,  because 
these  sides  lie  in  the  equipotential  surfaces,  and  therefore  no 
current  would  flow  along  these  metal  coatings.  Then  the  element 
of  volume  under  consideration  is  converted  into  a  small  plate 
condenser;  the  flux  density  and  the  stress  within  this  element  can 
be  considered  as  uniform,  so  that  formula  (277)  holds  true,  and 
we  have 

V  .......     (282) 


Differentials  are  used  because  both  the  volume  and  the  stored 
energy  are  infinitesimal.     The  density  of  energy 

W  =  dW/dV  =  !«G2,    .....     (283) 


and  has  the  same  expression  as  in  the  case  of  a  uniform  field;  but 
its  numerical  value  is  different  from  point  to  point,  because  G  is 
variable.  The  other  expressions  for  the  density  of  energy,  eqs. 
(278)  and  (279),  also  hold  true  for  the  points  of  a  non-uniformly 
stressed  dielectric,  provided  that  proper  values  of  D  and  G  are 
used  for  each  point. 

The  total  energy  stored  in  a  non-uniform  electrostatic  field  is 

W  =  \  fV  K&dV  =  \  CVGDdV  =  \  fV  D*dV/K;     (284) 
Jo  Jo  Jo 

two  more  expressions  may  be  written  in  which  I/a-  is  used  in  place 
of  K.  In  order  to  perform  the  integration  G  and  D  must  be  given 
as  functions  of  coordinates,  and  the  integration  extended  over 
the  whole  space  occupied  by  the  field.  Equations  (280)  and  (281) 
are  true  for  condensers  of  any  shape,  because  in  the  deduction  of 
these  formulae  no  assumption  is  made  as  to  the  particular  form  of 
the  dielectric  or  the  electrodes. 

The  expressions  for  the  electrostatic  energy  of  the  field, 
derived  above,  are  analogous  to  the  corresponding  ones  for  the 
potential  energy  of  stressed  elastic  bodies;  and  this  is  consistent 
with  the  assumed  behavior  of  dielectrics.  Consider  the  work 
necessary  per  cubic  centimeter  to  strain  mechanically  the  elastic 
fibers  of  a  given  material.  The  external  mechanical  force  being 
applied  gradually  (so  as  to  avoid  oscillations),  the  stress  varies 
from  zero  to  its  final  value  G.  Let  Gt  be  some  intermediate  value 
of  the  stress,  and  let  Dt  be  the  corresponding  strain.  The  same 
symbols  G  and  D  are  used  here  to  denote  the  mechanical  quantities 
analogous  to  electric  stress  and  displacement.  While  the  strain 


160  THE  ELECTRIC  CIRCUIT  [ART.  54 

increases  from  Dt  to  (Dt  +  dDt),  the  stress  Gt  may  be  considered 
constant;  the  infinitesimal  work  done  is  therefore  equal  to  Gt  dDt. 
The  total  work  of  deformation  is 

W 

But,  according  to  Hooke's  law  of  elasticity,  strains  are  propor- 
tional to  stresses,  so  that  a  linear  relation  exists  between  Dt  and 
Gt,  similar  to  eq.  (274).  We  thus  arrive  again  at  the  result  that 
the  work  necessary  to  strain  one  cubic  unit  of  an  elastic  material 
is  equal  to  |  <rD2. 

Prob.  Calculate  the  total  stored  energy,  and  the  density  of  energy,  in 
the  condenser  given  in  problem  2,  Art.  51. 

Ans.  20.52  milliwatt-seconds  (millijoules);  19.53  microjoules  per 
cubic  centimeter. 

54.  The  Permittance  and  Elastance  of  j  Irregular  Paths.1    In 

most  practical  cases  where  it  is  required  to  determine  the  per- 
mittance or  elastance  of  a  dielectric,  for  instance  in  high-tension 
apparatus,  the  geometric  shapes  of  the  metal  parts  and  of  the 
insulation  are  either  irregular  or  too  complicated  to  be  expressed 
analytically.  It  is  therefore  necessary  in  such  cases  to  determine 
the  shape  of  the  field  by  trials  and  approximations,  or  by  experi- 
ment. The  general  law,  substantiated  by  all  known  experiments, 
is  as  follows:  The  distribution  of  the  lines  of  force  and  equipotential 
surfaces  in  a  dielectric  is  such  as  to  make  the  total  permittance  a  maxi- 
mum, or  the  elastance  a  minimum. 

This  is  a  particular  case  of  the  general  law  of  nature  known  as 
the  law  of  minimum  resistance.  Let  a  condenser  of  irregular 
shape  (Fig.  48)  be  connected  to  a  source  of  unlimited  energy, 
having  a  constant  voltage  E.  The  law  of  minimum  resistance 
requires  that  the  dielectric  take  in  as  much  energy  as  is  compati- 
ble with  its  properties.  This  means  that  expression  (280)  must 
be  a  maximum ;  that  is,  with  constant  E,  the  permittance  C  must 
be  a  maximum,  or  the  elastance  S  a  minimum. 

Now  let  it  be  required  to  establish  a  given  flux  in  a  certain 
dielectric;  in  other  words,  let  Q  be  a  constant.  The  law  of  mini- 
mum resistance  requires  in  this  case  that  the  result  be  accom- 

1  The  treatment  is  similar  to  that  of  conductors  of  irregular  shape,  given 
in  Art.  10  of  this  book,  and  of  irregular  magnetic  paths  in  Art.  41  of  the 
Magnetic  Circuit. 


CHAP.  XV]  THE  DIELECTRIC  CIRCUIT  161 

plished  with  the  least  possible  expenditure  of  energy.  According 
to  eq.  (281),  we  have  again  the  same  condition  of  maximum  C  or 
minimum  S. 

Therefore,  in  order  to  calculate  the  permittance  (or  the  elas- 
tance)  of  a  given  dielectric,  or  to  find  the  flux  densities  and  stresses 
in  different  parts  of  it,  proceed  as  follows:  The  field  is  mapped 
out  into  small  cells  by  lines  of  force  and  equipotential  surfaces, 
drawing  them  to  the  best  of  one's  judgment;  the  total  permittance 
is  calculated  by  properly  combining  the  permittances  of  the  cells 
in  series  and  in  parallel.  Then  the  assumed  directions  are  some- 
what modified,  the  permittance  is  calculated  again,  and  so  on; 
until  by  successive  trials  the  positions  of  the  lines  of  force  are 
found  with  which  the  permittance  becomes  a  maximum. 

The  work  of  trials  is  made  more  systematic  by  following  a 
procedure  suggested  by  Lord  Rayleigh.  Imagine  infinitely  thin 
sheets  of  metal  (material  of  infinite  permittivity)  to  be  interposed 
at  intervals  into  the  field  under  consideration,  in  positions  approxi- 
mately coinciding  with  the  equipotential  surfaces.  If  these  sheets 
.exactly  coincided  with  the  actual  equipotential  surfaces,  the  total 
permittance  of  the  field  would  not  be  changed,  there  being  no 
tendency  for  the  flux  to  pass  along  the  equipotential  surfaces.  In 
any  other  position  of  the  conducting  sheets,  the  total  permittance 
of  the  field  is  evidently  increased.  Moreover,  these  sheets  become 
new  equipotential  surfaces  of  the  system,  because  no  difference  of 
potential  can  be  maintained  along  a  path  of  infinite  permittance. 
Thus,  by  drawing  in  the  given  field  a  system  of  surfaces  approxi- 
mately in  the  directions  of  the  true  equipotential  surfaces,  and 
assuming  these  arbitrary  surfaces  to  be  the  true  ones,  the  true 
elastance  of  the  path  is  reduced.  In  other  words,  by  calculating 
the  elastances  of  the  laminas  between  the  "  incorrect "  equipotential 
surfaces  and  adding  these  elastances  in  series,  one  obtains  an 
elastance  which  is  lower  than  the  true  elastance  of  the  field.  This 
gives  a  lower  limit  for  the  required  elastance  (or  an  upper  limit 
for  the  permittance)  of  the  field. 

Imagine  now  the  various  tubes  of  force  of  the  original  field 
wrapped  in  infinitely  thin  sheets  of  a  material  of  zero  permittivity 
or  infinite  elastivity  (absolute  insulator).  This  does  not  change 
the  elastance  of  the  paths,  because  no  flux  passes  between  the 
tubes.  But  if  these  wrappings  are  not  exactly  in  the  direction 
of  the  lines  of  force,  the  elastance  of  the  field  is  increased,  because 


162  THE  ELECTRIC  CIRCUIT  [Aim  54 

the  insulating  wrappings  displace  the  lines  of  force  from  their 
natural  positions.  Thus,  by  drawing  in  a  given  field  a  system  of 
surfaces  approximately  in  the  directions  of  the  lines  of  force, 
calculating  the  permittances  of  the  individual  tubes,  and  adding 
them  in  parallel,  an  elastance  is  obtained  which  is  higher  than  the 
true  elastance  of  the  field.  This  gives  an  upper  limit  for  the 
elastance  (or  a  lower  limit  for  the  permittance)  of  the  path  under 
consideration. 

Therefore,  the  practical  procedure  is  as  follows:  Divide  the 
field  to  the  best  of  your  judgment  into  cells,  by  equipotential 
surfaces  and  tubes  of  force,  and  calculate  the  elastance  of  the 
field  in  two  ways:  first,  by  adding  the  cells  in  parallel  and  the 
resultant  laminae  in  series;  secondly,  by  adding  the  cells  in  series 
and  the  resultant  tubes  in  parallel.  The  first  result  is  lower  than 
the  second.  Readjust  the  positions  of  the  lines  of  force  and  the 
equipotential  surfaces  until  the  two  results  are  sufficiently  close 
to  one  another;  an  average  of  the  last  two  results  gives  very 
nearly  the  true  elastance  of  the  field. 

One  difficulty  in  actually  following  out  the  foregoing  method, 
is  that  the  changes  in  the  assumed  directions  of  the  field,  that  will 
give  the  best  result,  are  not  always  obvious.  Dr.  Th.  Lehmann 
has  introduced  an  improvement  which  greatly  facilitates  the  lay- 
ing out  of  a  field.1  While  he  has  developed  his  method  for  the 
magnetic  field,  it  is  also  directly  applicable  to  the  electrostatic 
field.  We  shall  explain  this  method  as  applied  to  a  two-dimen- 
sional field,  though  theoretically  it  is  applicable  to  three-dimen- 
sional problems  also.  According  to  Lehmann,  lines  of  force  and 
equipotential  surfaces  are  drawn  at  such  distances  that  they 
inclose  cells  of  equal  elastance.  Consider  a  slice,  or  a  cell,  in  a 
two-dimensional  field,  a  centimeters  thick  in  the  third  dimension, 
and  of  such  a  form  that  the  average  length  I  of  the  cell  in  the 
direction  of  the  lines  of  force  is  equal  to  its  average  width  w  in  the 
perpendicular  direction.  The  elastance  of  such  a  cell  is  always 
equal  to  unity,  no  matter  whether  the  cell  itself  is  large  or  small. 
This  follows  from  the  fundamental  formula  for  elastance,  which 
in  this  case  becomes  S  =  al/(<r  X  w)  =  1. 

The  judgment  of  the  eye  helps  to  arrange  cells  of  widths 
equal  to  their  lengths,  in  proper  positions  with  respect  to  each 

1  "Graphische  Methode  zur  Bestimmung  des  Kraftlinienverlaufes  in  der 
Luft,"  Elektrotechnische  Zeitschrift,  Vol.  30  (1909),  p.  995. 


CHAP.  XV]  THE  DIELECTRIC  CIRCUIT  163 

other  and  to  the  electrodes;  the  next  approximation  is  apparent 
from  the  diagram,  by  observing  the  lack  of  equality  in  the  average 
width  and  length  of  the  cells.  Lord  Rayleigh's  condition  is 
secured  automatically,  since  the  combination  of  cells  of  equal 
elastance  leads  to  the  same  result,  whether  they  are  combined 
first  in  parallel  or  in  series.  After  a  few  trials  the  space  is  properly 
ruled,  and  it  simply  remains  to  count  the  number  of  cells  in  series 
and  in  parallel.  Dr.  Lehmann  shows  a  few  applications  of  his 
method  to  practical  cases  of  electrical  machinery,  and  the  reader 
is  referred  to  the  original  article  for  further  details. 

In  a  few  simple  cases,  as  for  instance  in  determining  the  elas- 
tance between  two  parallel  metallic  cylinders  of  circular  cross- 
section,  or  between  two  spheres,  the  principle  of  superposition  of 
electric  systems  in  equilibrium  can  be  used,  and  the  result  obtained 
without  trials.  This  principle  is  used  in  the  determination  of 
the  capacity  of  transmission  lines  and  cables,  in  the  next  two 
chapters.  In  two-dimensional  problems,  that  is,  in  determining 
the  shape  of  a  field  between  two  infinite  parallel  cylinders  of  any 
cross-sections  whatever,  the  properties  of  conjugate  functions  can 
also  be  used  in  some  simple  cases;  for  further  details  see  the 
references  in  Art.  10  above. 

Prob.  1.  Sketch  empirically  the  field  between  two  infinite  parallel 
cylinders  of  equal  circular  cross-section,  the  distance  between  the  centers 
being  a  few  times  larger  than  the  diameter.  Determine  the  lower  and 
upper  limits  of  permittance  per  unit  of  axial  length,  and  compare  the 
results  with  the  theoretical  formula  (320)  given  in  Art.  63  below. 

Prob.  2.  The  terminal  of  a  high-tension  transformer  consists  of  a 
long  vertical  rod  connected  to  the  winding,  and  a  torus  ring  concentric 
with  it,  connected  to  the  grounded  case.  The  ring  is  of  circular  cross- 
section,  and  is  placed  near  the  center  of  the  rod.  Assuming  the  insula- 
tion in  the  whole  field  to  be  of  the  same  permittivity,  calculate  by  trials 
the  elastance  of  the  combination,  with  certain  assumed  dimensions  of 
the  rod  and  the  ring. 

65.  The  Law  of  Flux  Refraction.  When  an  electrostatic 
flux  passes  from  one  dielectric  into  another  of  a  different  permit- 
tivity (Fig.  9,  Art.  11),  the  lines  of  force  suddenly  change  their 
direction  at  the  dividing  surface  A  B  between  the  media,  and 
in  so  doing  they  obey  the  law  of  refraction,  which  is 

tan  0!/tan  02  =  KI/KZ (285) 

Here  0i  and  02  are  the  angles  of  incidence  and  refraction  respec- 
tively, while  KI  and  KZ  are  the  permittivities  (relative  or  absolute) 


164  THE  ELECTRIC  CIRCUIT  [ART.  56 

of  the  two  media.  A  similar  law  is  proved  in  Art.  11  above  for 
the  electric  conducting  circuit,  and  in  Art.  41a  of  the  Magnetic 
Circuit  for  the  magnetic  flux.  The  proof  in  the  case  of  electro- 
static flux  is  similar  in  all  respects  to  that  given  in  Art.  11,  if  the 
student  will  use  the  words  flux  and  flux  density  in  place  of  current 
and  current  density,  and  permittivity  in  place  of  conductivity. 

Equation  (285)  shows  that  the  lower  the  permittivity  of  a 
dielectric  the  more  nearly  do  the  lines  of  force  in  it  approach  the 
direction  of  the  normal  NiNz  at  the  dividing  surface.  In  this 
way  the  path  of  a  displacement  between  two  given  points  is 
shortened  in  the  medium  of  lower,  and  lengthened  in  that  of 
higher  permittivity,  by  such  an  amount  in  each  case  that  the 
total  permittance  of  the  composite  condenser  is  larger  with 
refraction  than  without  it.  Hence,  the  existence  of  refraction  is 
a  necessary  consequence  of  the  general  law  of  least  resistance, 
mentioned  in  the  preceding  article. 

When  mapping  out  an  electrostatic  field  in  two  or  more 
media,  for  instance,  partly  in  a  solid  insulating  material,  partly 
in  oil,  and  partly  in  air,  the  lines  of  force  must  be  drawn  so  as  to 
satisfy  eq.  (285)  at  the  dividing  surfaces.  The  permittance  of 
the  part  of  the  circuit  in  any  one  of  the  media  will  not  be  a 
maximum,  although  the  permittance  of  the  whole  combination 
must  be  a  maximum.  It  will  thus  be  seen  that  the  problem, 
while  quite  simple  in  theory,  is  by  no  means  an  easy  one  in  numer- 
ical applications,  especially  with  the  shapes  of  surfaces  used  in 
the  construction  of  commercial  high-tension  apparatus.  It  is 
advisable  for  the  student  to  train  his  eye  in  sketching  lines  of 
force  in  adjoining  media  of  different  permittivities,  conforming 
the  field  at  least  approximately  to  eq.  (285).  This  can  be  con- 
veniently done  on  available  drawings  of  high-tension  transformers, 
switches,  lightning  arresters,  etc.1 

56.  The  Dielectric  Strength  of  Insulating  Materials.  The 
proportionality  between  stress  and  flux  density,  indicated  by 
eqs.  (273)  and  (274),  holds  only  up  to  a  certain  limit;  in  this 
respect  it  is  similar  to  the  proportionality  between  stresses  and 
strains  in  an  elastic  body.  After  a  certain  limit  of  dielectric 
flux  density  or  of  voltage  gradient  has  been  exceeded,  the  material 

1  See  also  some  interesting  sketches  and  experiments  in  Professor  W.  S. 
Franklin's  article  on  "  Dielectric  Stresses  from  the  Mechanical  Point  of  View," 
in  the  General  Electric  Review,  Vol.  14,  June,  1911. 


CHAP.  XV] 


THE  DIELECTRIC  CIRCUIT 


165 


weakens  and  finally  breaks  down.  The  phenomena  of  failure  of 
electric  insulation  and  the  subsequent  disruptive  discharge  are 
too  well  known  to  need  a  description  here. 

The  values  of  the  critical  voltage  gradient  Gmax  and  of  the 
corresponding  flux  density  Dmax,  at  which  some  of  the  more  im- 
portant materials  break  down,  are  given  in  the  last  two  columns 
of  the  table  below.  In  designing  insulation,  the  stresses  must 
be  kept  well  below  these  critical  values,  the  factor  of  safety  de- 
pending upon  the  importance  of  the  apparatus,  possibility  of  over- 
potentials,  and  the  gradual  deterioration  of  the  insulation  by  heat, 
chemical  action,  moisture,  and  so  forth.  The  values  in  the  table 
are  principally  intended  to  give  the  student  an  idea  of  the  order 
of  magnitude  of  Gmax  and  Dmax.  More  accurate  data  will  be  found 
in  electrical  handbooks  and  pocket  books;  in  important  cases  these 
design  constants  should  be  based  upon  test  data  obtained  on  the 
material  in  hand. 


Substance. 

Relati  ve  permit- 
tivity ,  or  specific 
inductive  capac- 
ity K. 

Rupturing  volt- 
age gradient, 
Gmax  inkv. 
per  mm. 

Rupturing  values  of 
dielectric  flux  den- 
sity, DmaXt  in  me. 
per  sq.  cm. 

Air 

1 

3 

0  00265 

Glass,  different  kinds  
Mica,  natural  and  built  up..  .  . 
Porcelain 

3-8 
5-8 
4  4 

9-11 
24-40 
13-22 

0.024  -0.078 
0.110  -0.280 
0  050  -0  085 

Rubber,  pure  
Rubber,  vulcanized  
Transformer  oil  

2.2 
2.7 
2-2.2 

28-35 
21-28 
14 

0.055  -0.068 
0.050  -0.067 
0.025  -0.027 

Vacuum  

0.99 

It  will  be  seen  from  the  second  column  of  the  table  that  the 
permittivities  of  solid  and  liquid  dielectrics  are  larger  than  that 
of  air;  in  other  words,  they  are  more  yielding  to  electric  stress 
than  the  air.  This  does  not  mean,  however,  that  they  break 
down  at  a  lower  voltage  gradient  than  the  air.  On  the  contrary, 
the  third  and  fourth  columns  show  that  the  dielectrics  commonly 
used  in  electrical  engineering  are  considerably  stronger  electrically 
than  the  air,  in  that  they  can  stand  several  times  the  electric  stress 
and  displacement  at  which  the  air  breaks  down. 

There  does  not  seem  to  be  any  relation  between  the  values  of 
elastivity  and  critical  voltage  gradient.  One  indicates  the  elec- 
trical elasticity  of  the  material,  the  other  its  ultimate  strength. 
They  are  analogous  to  the  modulus  of  elasticity  and  the  rupturing 
stress  respectively  in  the  mechanics  of  materials.  Air,  from  an 


166  THE  ELECTRIC  CIRCUIT  [AKT.  57 

electrical  point  of  view,  may  be  compared  to  a  material  of  great 
stiffness,  but  one  which  breaks  at  a  comparatively  small  elonga- 
tion. On  the  contrary,  mica  may  be  likened  to  a  material  which 
is  comparatively  yielding,  but  can  stand  a  very  large  elongation 
before  it  is  ruptured;  so  that,  in  spite  of  a  smaller  elastivity, 
a  much  higher  stress  is  required  to  rupture  mica  than  air.  The 
student  is  advised  to  make  clear  to  himself  these  two  separate 
properties  of  dielectrics.  A  rational  design  of  high-tension 
insulation  depends  essentially  upon  a  distinct  understanding  of 
them. 

Dielectric  strength  may  be  properly  given  as  the  critical  flux 
density,  Dmnz,  but  for  practical  purposes  it  is  more  convenient  to 
express  it  as  the  critical  voltage  gradient,  Gmax,  at  which  the 
dielectric  is  broken  down.  When  a  dielectric  is  used  for  insula- 
tion in  the  form  of  thin  sheets  having  a  comparatively  large 
radius  of  curvature,  the  flux  density,  and,  consequently,  the  volt- 
age gradient,  are  practically  uniform  throughout,  so  that  Gmnx  = 
Gave.  When,  however,  the  layer  of  dielectric  is  thick  as  compared 
to  its  radius  of  curvature,  as  for  instance  in  the  insulation  of 
high-tension  machines,  or  when  air  or  oil  are  tested  between 
two  spherical  terminals,  the  use  of  the  average  voltage  gradient 
Gave  =  E/l  leads  to  wrong  results.  The  only  proper  way  in 
this  case  is  to  calculate  the  voltage  gradient  for  the  place  where 
it  is  a  maximum,  and  to  see  that  it  does  not  exceed  the  critical 
value  determined  from  previous  tests.  A  breakdown  in  one 
point  of  the  dielectric  results  in  an  increase  of  gradient  in  others, 
and  possibly  in  a  complete  failure. 

Prob.  1.  Show  how  the  values  in  the  last  column  of  the  table  are 
derived  from  those  in  the  two  preceding  columns. 

Ans.    Dmax  =  0.08842  KGmnx  X  1Q-*. 

Prob.  2.  A  certain  material  stood  about  82  kv.  in  a  layer  3.7  mm. 
thick.  What  voltage  gradient  can  be  allowed  in  this  material  at  a  factor 
of  safety  of  2?  Ans.  11  kv.  per  mm. 

Prob.  3.  Assuming  the  relative  permittivity  of  the  insulation  in  the 
preceding  problem  to  be  2.5,  what  is  the  density  of  energy  at  which  the 
material  is  broken  down? 

Ans.     5.45  X  10~3  joules  per  cubic  centimeter. 

67.  The  Electrostatic  Corona.  The  phenomena  which  ac- 
company the  electrical  breaking-down  of  air  deserve  special 
mention  in  view  of  their  great  practical  importance.  When  the 
voltage  at  the  terminals  of  an  air  condenser  is  raised  sufficiently 


CHAP.  XV]  THE  DIELECTRIC  CIRCUIT  167 

high,  a  pale  violet  light  appears  at  the  edges,  at  the  sharp  points, 
and  in  general  at  the  protruding  parts  having  a  comparatively 
small  radius  of  curvature.  This  silent  discharge  into  air,  due  to 
an  excessive  electrostatic  flux  density,  is  called  the  electrostatic 
corona.  In  the  regions  where  the  corona  appears,  the  air  is  elec- 
trically "  broken  down  "  and  ionized,  so  that  it  becomes  a  con- 
ductor of  electricity.  When  the  voltage  is  raised  still  higher  the 
so-called  brush  discharge  takes  place,  until  the  whole  thickness 
of  the  dielectric  is  broken  down,  and  a  disruptive  discharge,  or 
spark,  jumps  from  one  electrode  to  the  other. 

When  the  electrodes  have  projecting  parts  or  sharp  edges,  the 
corona  is  formed  at  a  voltage  far  below  that  at  which  the  disrup- 
tive discharge  occurs;  the  operating  voltage  of  such  devices  is 
generally  limited  to  that  at  which  the  corona  forms.  No  corona 
is  usually  permissible  in  regular  operation;  first,  because  it  may 
involve  an  appreciable  loss  of  power;  secondly,  because  the  dis- 
charge, if  allowed  to  play  on  some  other  insulation,  will  soon  char 
and  destroy  it.  There  are  cases,  however,  in  which  some  corona 
formation  is  harmless.  The  air  which  is  broken  down  becomes  a 
part  of  the  electrode,  smoothes  down  the  shape  of  the  protruding 
metallic  parts,  increases  their  area,  and  thus  reduces  the  danger- 
ous flux  density  and  makes  it  more  uniform.  It  is  of  advantage 
to  operate  certain  parts  of  a  very  high-tension  line  at  nearly  the 
critical  voltage.  Any  voltage  rise  on  the  line  due  to  lightning 
or  surges  is  automatically  relieved  by  a  corona  loss  into  the 
atmosphere;  so  that  the  line  may  be  made  self -protected,  without 
lightning  arresters. 

The  formation  of  corona  must  be  kept  in  mind  in  the  design 
of  high-tension  insulation,  and  in  high-potential  tests.  Shapes 
and  combinations  of  parts  which  lead  to  high  or  non-uniform 
dielectric  flux  densities  should  be  avoided.  Fig.  48  shows  the 
reason  why  the  dielectric  flux  density,  or  the  potential  gradient, 
is  higher  near  protruding  parts.  The  equipotential  surfaces,  for 
obvious  geometrical  reasons,  lie  closer  to  each  other  near  such 
parts,  while  at  a  reasonable  distance  from  the  electrodes  the 
shape  of  the  equipotential  surfaces  is  not  affected  by  small  irregu- 
larities in  the  shape  of  the  metallic  parts. 

It  will  be  seen  from  the  table  in  the  preceding  article  that  the 
air  is  broken  down  when  the  voltage  gradient  exceeds  3000  volts 
per  millimeter.  Let  this  be  the  case  at  the  point  P  (Fig.  48). 


168  THE  ELECTRIC  CIRCUIT  [Aux.  57 

The  voltage  gradient  has  this  value  only  at  the  very  surface  of 
the  conductor,  because  the  lines  of  force  immediately  spread  out 
in  the  air.  Thus,  only  a  very  small  portion  of  the  air  is  broken 
down  and  becomes  part  of  the  conducting  electrode.  No  visual 
corona  is  formed,  however.  Let  now  the  voltage  be  raised  still 
further;  then  the  next  layer  of  air  is  broken  down  and  becomes 
part  of  the  electrode.  When  a  sufficiently  thick  layer  of  air  is  thus 
ionized,  a  visual  corona  is  formed  around  the  point  P.  Consider- 
ing the  actual  surface  of  the  metal  as  the  starting  point,  the  volt- 
age gradient  at  that  point  now  would  seem  to  be  higher  than  3000 
volts  per  millimeter.  This  higher  value  is  called  the  visual  voltage 
gradient  as  distinguished  from  the  disruptive  voltage  gradient  of 
3000.  The  student  should  not  be  misled  by  these  names.  In 
reality  the  voltage  gradient  does  not  exceed  3000,  because  beyond 
this  the  air  becomes  part  of  the  electrode;  however,  the  concept  of 
visual  voltage  gradient  is  convenient  in  calculations. 

In  reality  the  phenomenon  of  ionization  of  air  and  formation 
of  the  corona  is  not  as  simple  as  described  above,  especially  around 
conductors  of  small  diameter,  say  less  than  6  mm.  The  physical 
state  of  the  layer  of  air  adjacent  to  the  conductor  seems  to  be  in 
some  peculiar  way  affected  by  it,  and  the  critical  voltage  gradient 
apparently  depends  in  this  case  upon  the  diameter  of  the  conductor. 
A  discussion  of  numerical  values  and  of  physical  theories  is  out- 
side the  scope  of  this  book;  and  the  student  is  referred  for  infor- 
mation to  the  numerous  articles  on  the  subject  that  appear  in  the 
leading  periodicals,  and  in  the  transactions  of  the  electrical  engi- 
neering societies  in  this  country  and  abroad.1 

Quite  extensive  tests  on  corona  formation,  critical  voltage,  and 
the  accompanying  loss  of  power  were  performed  by  the  General 
Electric  Company,  in  1910-11,  and  have  been  described  by 
Mr.  Peek.2  The  student  is  referred  to  his  article  for  numerical 
data;  the  results  are  given  on  the  first  few  pages  of  the  article, 
and  are  illustrated  by  a  numerical  example. 

1  See,  for  instance,  H.  J.  Ryan,  "  Open  Atmosphere  and  Dry  Transformer 
Oil  as  High-voltage  lusulators,"  in  the  Trans.  Amer.  Inst.  Electr.  Engrs., 
Vol.  30,  Jan.,  1911.  This  paper  is  a  splendid  exposition  of  the  subject  by 
one  of  the  pioneer  investigators  of  the  corona,  and  contains  numerous  refer- 
ences to  other  articles  on  the  subject.  Professor  J.  B.  Whitehead's  experi- 
mental investigations  are  particularly  noteworthy. 

1  F.  W.  Peek,  Jr.,  "The  Law  of  Corona  and  the  Dielectric  Strength  of 
Air,"  Trans.  Amer.  Inst.  Electr.  Engrs.,  Vol.  30,  July,  1911;  also  Vol.  31. 


CHAP.  XV]  THE  DIELECTRIC  CIRCUIT  169 

Prob.  1.  Assuming  that  under  certain  conditions  a  corona  is  formed 
when  the  dielectric  flux  density  exceeds  0.0034  microcoulombs  per  square 
centimeter,  calculate  the  factor  of  safety  of  a  25-cycle  transmission  line 
for  which  the  charging  current  is  0.12  amp.  per  kilometer,  the  diameter 
of  the  conductors  being  12  mm.  Solution:  The  line  is  charged  during 
0.01  of  a  second,  and  the  average  charging  current  is  0.12/1.11  =  0.108 
amp.;  hence,  the  maximum  electrostatic  displacement  in  the  air  is  1080 
microcoulombs  per  kilometer.  The  surface  of  each  conductor  is  377,000 
sq.  cm.  per  kilometer,  so  that  the  density  of  displacement  is  1080/377,000 
=  0.002865  microcoulombs  per  square  centimeter,  and  the  factor  of 
safety  is  34/28.65  =  1.20. 

58.  Dielectric  Hysteresis  and  Conductance.  When  an  al- 
ternating voltage  is  applied  at  the  terminals  of  a  condenser,  the 
dielectric  is  subjected  to  periodic  stresses  and  displacements. 
If  the  material  were  perfectly  elastic,  no  energy  would  be  lost 
during  one  complete  cycle,  because  the  energy  stored  during  the 
periods  of  increase  in  voltage  would  be  given  up  to  the  circuit  when 
the  voltage  decreased.  In  reality,  the  electric  elasticity  of  solid 
and  liquid  dielectrics  is  not  perfect,  so  that  the  applied  voltage 
has  to  overcome  some  kind  of  molecular  friction,  in  addition  to 
the  elastic  forces.  The  work  done  against  friction  is  converted 
into  heat,  and  is  lost,  as  far  as  the  circuit  is  concerned.  The 
phenomenon  is  similar  to  the  familiar  magnetic  hysteresis,  and  is 
therefore  called  dielectric  hysteresis.  The  energy  lost  per  cycle 
is  proportional  to  the  square  of  the  applied  voltage,  because  both 
the  displacement  and  the  stress  are  proportional  to  the  voltage. 

When  stresses  are  well  below  the  ultimate  limit  of  the  material, 
the  loss  of  power  caused  by  dielectric  hysteresis  is  exceedingly 
small.  Some  investigators  are  even  in  doubt  as  to  whether  it 
exists  at  all.  There  is  often  an  appreciable  loss  of  power  in  com- 
merical  condensers,  but  this  loss  can  be  mostly  attributed  to  the 
fact  that  dielectrics  are  not  perfect  insulators.  While  their  ohmic 
resistance  is  exceedingly  high,  as  compared  with  that  of  metals, 
they  nevertheless  conduct  some  current,  especially  at  high  voltages. 
Thus,  the  observed  loss  of  power  and  the  heating  of  condensers 
may  be  simply  ascribed  to  the  PR  loss  in  the  insulation.  More- 
over, small  coronas  can  form  at  the  edges  and  projecting  parts, 
even  at  the  operating  voltage,  and  thus  be  an  additional  source 
of  loss.  Some  small  loss  is  also  due  to  the  ohmic  resistance  and 
eddy  currents  in  the  metallic  sheets  which  compose  the  electrodes 
or  plates  of  the  condenser. 


170  THE  ELECTRIC  CIRCUIT  [ART.  58 

An  imperfect  condenser,  that  is,  one  which  shows  a  loss  of 
power  from  one  cause  or  another,  can  be  replaced  for  purposes 
of  calculation  by  a  perfect  condenser  with  an  ohmic  conductance 
.shunted  around  it.  This  conductance,  or  "  leakance,"  as  some 
authors  call  it,  is  selected  of  such  a  value  that  the  PR  loss  in  it  is 
equal  to  the  loss  of  power  from  all  causes  in  the  given  imperfect 
condenser.  The  actual  current  through  the  imperfect  condenser 
is  considered  then  as  consisting  of  two  components,  —  the  leading 
reactive  component  through  the  ideal  condenser,  and  the  loss 
component,  in  phase  with  the  voltage,  through  the  shunted  con- 
ductance. In  this  way,  imperfect  condensers  can  be  treated  graphi- 
cally or  analytically,  according  to  the  ordinary  laws  of  the  electric 
circuit. 

Prob.  1.  A  certain  kind  of  condenser  shows  a  loss  of  power  of  about 
17.9  watts  per  microfarad,  at  2200  volts,  25  cycles.  By  what  fictitious 
conductance  should  an  ideal  condenser  be  shunted,  in  order  to  replace 
a  condenser  of  this  kind  having  a  capacity  of  1.5  mf,? 

Ans.    5.55  rnicromhos. 


CHAPTER  XVI 

ELASTANCE  AND  PERMITTANCE  OF  SINGLE-PHASE 
CABLES  AND  TRANSMISSION  LINES 

59.  The  Elastance  of  a  Single-core  Cable.     A  cross-section 
of  a  single-core  cable  is  shown  in  Fig.  49.     The  round  conductor 
in  the  center  is  assumed  to  be  solid  (not 
stranded)  for  the  sake  of  simplicity.     It  is 
surrounded  by  a  layer  of  insulation,  and  is 
protected  on  the  outside  by  a  lead  sheath- 
ing.    Let  such  a  cable  be  subjected  to  a 
difference  of  potential  between  the  core  and 
the  sheathing;  for  instance,  let  one  pole  of 
a  battery  be  connected  to  the  core  and  the 
other  pole  to  the  sheathing.     Let  it  be  re- 
quired to  find  the  permittance  or  the  elast-  ^  fsin^core  OTlon" 
ance  of  the  dielectric   for  a   certain   axial      centric  cable, 
length  I  of  the  cable. 

For  reasons  of  symmetry,  the  lines  of  force  are  radial  straight 
lines  between  the  two  metal  surfaces,  and  the  equipotential  sur- 
faces are  concentric  cylinders.  Consider  the  insulation  to  be  sub- 
divided into  concentric  layers  of  infinitesimal  thickness.  The 
elastances  of  these  layers  are  all  in  series,  so  that  it  is  sufficient  to 
express  analytically  the  elastance  of  a  layer  having  a  radius  x 
and  thickness  dx,  and  to  integrate  this  expression  between  the 
limits  a  and  6,  where  a  is  the  radius  of  the  core,  and  6  is  that  of 
the  inner  surface  of  the  sheathing.  The  elastance  of  the  layer  in 
question  is 

dS  =  <rdx/(2irxl), (286) 

dx  and  2  irxl  being  respectively  the  length  and  cross-section  of  the 
path  of  the  radial  flux.  Integrating  this  expression  between  the 
limits  a  and  6  gives 

S=  (<r/2irZ)Ln(6/a), (287) 

the  abbreviation  Ln  standing  for  natural  logarithm. 

171 


172  THE  ELECTRIC  CIRCUIT  [ART.  59 

For  practical  calculations  it  is  convenient  to  modify  this  formula 
in  three  respects;  namely,  (a)  to  introduce  the  relative  permit- 
tivity K  of  the  insulating  material,  (b)  to  express  I  in  kilo- 
meters, and  (c)  to  use  common  logarithms.  Making  these 
changes,  we  finally  obtain 

S  =  C~1  =  (41.45/XZ)  log  (b/a)  megadarafs. .     .     (288) 

For  the  permittance  (capacity)  per  kilometer  we  have  accordingly 

C'  =  C/l  =  0.0241  K/\og(b/a)  microfarads  per  kilometer.1    (289) 

In  some  cases  it  is  necessary  to  know  the  voltage  across  a 
certain  part  of  the  insulation,  for  instance  between  the  radii  r  and 
r'.  Applying  formula  (287)  to  this  case,  for  I  =  1  cm.,  we  get 
S'rr  =  (ff/2v)  Ln  (r'/r).  The  voltage  drop  Err>  from  r  to  r'  is  equal 
to  this  elastance  multiplied  by  the  electric  displacement  Q'  per 
centimeter  length  of  the  cable.  Or 

Err'  =  Sfrr'  -  Q'  =  (ffQr /2  TT)  Ln  (r'/r}.      .     .     (290) 

This  formula  finds  its  important  application  below  in  the  calcula- 
tion of  the  permittance  of  single-phase  and  polyphase  transmission 
lines.  It  is  absolutely  essential  to  agree  in  regard  to  the  signs  in 
eq.  (290).  In  the  applications  that  follow,  Q'  is  taken  with  the 
plus  sign  when  the  positive  displacement  is  directed  from  the  con- 
ductor, and  with  the  minus  sign  when  it  is  directed  towards  the 
conductor.  It  is  also  important  to  write  the  distances  r'  and  r  in 
the  order  given,  because  interchanging  r'  and  r  in  eq.  (290) 
changes  the  sign  of  Err'. 

If  the  insulation  consists  of  two  or  more  concentric  layers  of 
different  materials,  the  elastances  of  the  layers  are  calculated 
separately,  according  to  formula  (288),  and  then  added  in  series. 
The  permittance  of  the  cable  as  a  whole  is  the  reciprocal  of  this 
resultant  .elastance.  The  same  formulae  apply  to  a  concentric 
cable  without  sheathing,  the  outside  conductor  taking  the  place 
of  the  sheathing  as  far  as  stresses  in  the  dielectric  are  concerned. 
With  two  cylindrical  conductors  side  by  side  the  elastance  is  cal- 
culated as  shown  in  Art.  63  below.  With  three  conductors  the 
theory  is  rather  difficult;  as  is  also  the  case  when  the  conduc- 
tors are  not  of  circular  cross-section.  Those  interested  will  find 

1  This  simple  derivation  of  the  formula  for  the  capacity  of  a  single-core 
cable  demonstrates  in  a  particularly  striking  manner  the  usefulness  of  the 
concept  of  elastance. 


CHAP.  XVI]       ELASTANCE  OF  CABLES  AND  LINES  173 

extensive  literature  on  the  subject  in  the  European  electrical 
magazines  and  proceedings  of  electrical  societies.  In  practice, 
the  permittance  of  such  cables  is  usually  determined  by  test. 

The  distribution  of  the  electric  stresses  in  a  single-core  cable 
is  of  considerable  practical  importance.  The  total  displacement 
Q  being  the  same  through  every  concentric  layer  of  the  dielectric, 
the  flux  density  and  consequently  the  stress  is  a  maximum  at  the 
surface  of  the  inner  core.  For  a  layer  of  radius  x  we  have 

Q  =  Dx'2-n-xl  =  const., (291) 

where  Dx  is  the  density  of  displacement  through  that  layer. 
Hence, 

Dxx  =  const., (292) 

which  means  that  the  density  of  displacement  is  inversely  pro- 
portional to  the  distance  from  the  center.  Since  displacements 
are  proportional  to  stresses  (with  a  uniform  insulation),  we  also 
have 

Gxx  =  const (293) 

A  useful  relation  between  the  total  applied  voltage  E  and  the 
stress  Gx  at  a  given  point  in  the  dielectric  can  be  deduced  from 
eq.  (293).  We  have 

Gx  =  const,  /x; 

and  if  we  multiply  both  sides  by  dx  and  integrate  between  a  and 
b,  remembering  that  voltage  is  the  line  integral  of  intensity,  we 
obtain 

Gxdx  =  E  =  (const.)  Ln  (6/a). 

Eliminating  the  constant  between  these  two  equations,  gives 

Gx  =  E/[x  Ln  (6/a)] (294) 

Equations  (292)  and  (293)  show  that  a  homogeneous  dielectric 
is  fully  utilized  with  regard  to  its  dielectric  strength  only  at  the 
surface  of  the  core,  the  stress  gradually  decreasing  toward  the 
periphery.  This  condition  could  be  helped  by  gradually  increasing 
the  elastivity  of  the  material  toward  the  sheathing,  so  as  to  in- 
crease the  voltage  drop  and  the  stresses  there.  If  the  elastivity 
of  each  layer  could  be  made  exactly  proportional  to  its  radius, 
the  stress  Gx  would  be  the  same  throughout  the  dielectric.  Such 
a  condition  would  be  an  ideal  one,  with  regard  to  economy  in 


174  THE  ELECTRIC  CIRCUIT  [ART.  59 

material,  provided  that^  the  dielectric  strength  of  the  "  variable 
insulation  "  were  constant. 

This  purely  theoretical  conclusion  leads  to  the  important 
practical  question  of  the  grading  of  insulation  of  cables.  With 
high-tension  cables,  in  which  the  thickness  of  insulation  is  large, 
it  pays  to  provide  two  or  more  layers  of  different  materials,  utiliz- 
ing their  permittivities  and  ultimate  strengths  in  the  most  ad- 
vantageous manner.  The  problem  is  primarily  to  relieve]  the 
stress  near  the  inner  core,  and  this  is  done  by  placing  near  it  a 
layer  of  insulation  of  high  permittivity,  so  as  to  cause  a  low  volt- 
age drop.  One  case  where  the  opposite  arrangement  would  be 
advantageous  is  in  a  low-voltage  cable  in  which  it  is  desired  to 
keep  the  total  permittance  as  low  as  possible  (for  example,  to 
obtain  small  capacity  or  low  charging  current  at  high  frequen- 
cies). In  this  case  the  layer  surrounding  the  core  must  have  as 
high  an  elastance  as  possible,  because  it  is  this  layer  that  contrib- 
utes most  to  the  total  elastance  of  the  cable.  With  a  clear 
understanding  of  these  principles,  the  student  will  be  able  to 
design  a  graded  insulation  for  given  conditions,  if  he  knows  the 
properties  of  the  available  materials.1 

Prob.  1.  A  single-core  cable  receives  a  charge  of  1.18  millicoulombs 
per  kilometer  when  a  continuous  voltage  of  12  kv.  is  applied  between 
the  core  and  the  sheathing.  The  core  consists  of  a  solid  conductor  the 
diameter  of  which  is  5  mm. ;  the  insulation  is  9.5  mm.  thick.  Determine 
the  value  of  the  relative  permittivity  of  the  material  of  insulation,  and 
the  extreme  values  of  the  dielectric  flux  density. 

Ans.  K  =  2.78;  Dmaz  =  0.00750;  Dmin  =  0.00156  microcoulombs 
per  sq.  cm. 

Prob.  2.  The  insulation  used  in  the  cable  specified  in  the  preceding 
problem  breaks  down  at  a  flux  density  of  0.062  me.  per  sq.  cm.  Show 
that  the  critical  voltage  for  the  cable  is  about  70  alternating  kilovolts. 

Prob.  3.  What  is  the  ratio  between  the  maximum  and  the  average 
stress  in  the  insulation  in  problem  1?  Ans.  2.42.2 

Prob.  4.  Deduce  formula  (290)  from  the  fact  that  the  voltage  is  the 
line  integral  of  the  electric  intensity. 

1  For  a  theoretical  treatment  of  the  grading  of  insulation,  and  for  the 
bibliography  of  the  subject,  see  H.  S.  Osborne,  Potential  Stresses  in  Die- 
lectrics (1910),  a  thesis  presented  to  the  Massachusetts  Institute  of  Tech- 
nology for  the  degree  of  Doctor  of  Engineering. 

*  There  is  a  tendency  in  practice  to  deal  with  average  stresses  even  when 
the  field  is  far  from  being  uniform.  The  answer  to  this  problem  shows  that 
one  has  to  be  careful  in  using  an  average  value,  unless  its  ratio  to  the  maxi- 
mum stress  is  known. 


CHAP.  XVI]       ELASTANCE  OF  CABLES  AND  LINES  175 

Prob.  6.  Show  by  actual  calculation  that  in  the  foregoing  cable  the 
maximum  stress  in  the  dielectric  is  reduced  by  increasing  the  diameter  of 
the  conductor  to  7.5  mm.,  with  the  same  diameter  of  the  sheathing. 
This  is  in  spite  of  the  fact  that  the  insulation  becomes  thinner,  and 
consequently  the  average  stress  greater. 

Prob.  6.  Referring  to  the  preceding  problem,  show  that  it  is  of  ad- 
vantage to  make  the  ratio  b/a  about  equal  to  «,  where  e  =  2.71828  .  .  . 
is  the  base  of  the  natural  system  of  logarithms.  If  the  diameter  of  the 
conductor  be  further  increased,  so  that  the  ratio  b/a  becomes  less  than 
e,  the  maximum  stress  does  not  continue  to  decrease,  but  increases  in- 
stead. Solution:  The  stress  at  the  core  is  Ga  =  E/[aLn  (b/a)]  according 
to  eq.  (294) .  As  a  varies,  Ga  reaches  its  maximum  when  dGa/da  =  0. 
Differentiating,  we  get 

dGa/da  =  E  [1  -  Ln  (6/a)]/[a  Ln  (b/a)]2  =  0; 
whence,  1  —  Ln  (6/a)  =  0,  or  b/a  =  t. 

Prob.  7.  Explain  the  following  deduction  from  the  theorem  stated  in 
the  preceding  problem.  In  a  concentric  cable  subjected  to  an  excessive 
voltage,  if  the  insulation  is  quite  thick,  the  layer  around  the  inner  core  is 
first  gradually  destroyed  or  charred  up  to  a  certain  thickness,  and  then 
the  rest  of  the  insulation  suddenly  breaks  down.  With  a  thin  layer  of 
insulation  no  such  phenomenon  is  observed. 

Prob.  8.  A  cable  is  provided  with  several  concentric  layers  of  insula- 
tion, the  external  radii  of  which  are  61,  62,  etc.,  and  the  relative  permit- 
tivities, Ki,  K2,  etc.  Show  that  the  elastance  of  the  cable  is  expressed 
by  the  formula 

S  =  (41.45/0  [Krl  log  (bi/a)  +  Krl  log  (62/6,)  +  Krl  log  (63/62)  +  etc.]. 

Prob.  9.  Show  that  in  a  single-core  cable  the  density  of  energy  stored 
in  the  dielectric  varies  inversely  as  the  square  of  the  distance  from  the 
center. 

Prob.  10.  A  conductor  2  a  cm.  in  diameter  is  surrounded  by  a  con- 
centric metal  cylinder  of  26  cm.  inside  diameter.  What  alternating  volt- 
age can  be  allowed  between  the  cylinder  and  the  conductor  at  a  factor  of 
safety  k  against  the  formation  of  corona? 

Ans.  E  =  18.4  a  (Dc  X  103/&)  log  (6/a)  effective  kilovolts,  where  Dc 
is  the  flux  density  in  microcoulombs  per  sq.  cm.,  at  which  corona  is 
formed. 

Prob.  11.  Show  that  the  elastance  of  the  dielectric  between  two  con- 
centric spheres  of  radii  a  and  6  is  equal  to  (o-a/4  *-./?)  (I/a  —  1/6)  mega- 
darafs. 

Prob.  12.  Show  that  with  two  concentric  spheres  the  equation  corre- 
sponding to  (294)  is  Gx  =  E/[x*  (a~l  -  &-»)]. 

Prob.  13.  Apply  the  formulae  given  in  the  text  above  to  the  theory  of 
a  condenser-type  terminal.1 

1  See  A.  B.  Reynders,  "Condenser  Type  of  Insulation  for  High-tension 
Terminals,"  Trans.  Amer.  Inst.  Electr.  Engrs.,  Vol.  28  (1909),  p.  209. 


176  THE  ELECTRIC  CIRCUIT  [ABT.  60 

60.  The  Elastance  of  a  Single-phase  Line.  The  general 
character  of  the  electrostatic  field  between  two  infinite  parallel 
conductors  is  shown  in  Fig.  50.  The  lines  of  force  are  arcs  of 
circles  extending  from  one  metal  surface  to  the  other;  the  equi- 
potential  surfaces  are  circular  cylinders  eccentric  with  respect  to 
the  conductors  (see  Art.  62  below).  It  is  required  to  calculate 
the  elastance  of  the  air  between  the  two  conductors,  for  a  unit 
axial  length  of  the  line.  Knowing  this  elastance,  the  charging 
current  of  the  line  can  be  calculated  for  a  given  frequency.  This 
elastance,  or  its  reciprocal,  the  permittance,  is  used  in  the  pre- 
determination of  the  regulation  of  a  transmission  line  (Art.  68 
below). 

We  shall  consider  in  this  article  the  usual  practical  case  in 
which  the  radius  a  of  the  conductors  is  small  as  compared  with 
the  interaxial  distance  b.  It  is  shown  in  Art.  63  below  how  to 
determine  the  elastance  when  the  diameters  of  the  cylinders  are 
comparatively  large. 

For  purposes  of  analysis  it  is  convenient  to  consider  the  field 
shown  in  Fig.  50  as  the  result  of  the  superposition  of  two  simple 
radial  fields  similar  to  that  in  Fig.  49.  Consider  the  conductor 
A ,  together  with  a  concentric  cylinder  of  an  infinitely  large  radius, 
as  one  electric  system.  Let  the  conductor  B  with  a  similar  con- 
centric cylinder  form  another  independent  system.  Let  the  con- 
ductor A  be  connected  to  the  positive  pole  of  a  battery  of  voltage 
E,  the  conductor  B  to  the  negative  pole,  and  the  two  cylinders  at 
infinity  to  the  middle  point  of  the  battery.  In  the  first  concentric 
condenser  the  displacement  of  positive  electricity  is  from  the  con- 
ductor A  to  the  infinite  cylinder,  while  in  the  second  system  the 
positive  displacement  is  from  the  infinite  cylinder  toward  the  con- 
ductor B.  The  displacements  due  to  the  two  systems  are  equal 
and  opposite  at  the  two  infinite  cylinders,  and  the  cylinders  them- 
selves coincide  at  infinity,  because  the  distance  AB  between  their 
axes  is  infinitely  small  as  compared  with  their  radii.  Hence,  the  two 
displacements  at  the  cylinders  cancel  each  other,  and  the  combina- 
tion of  the  two  cylindrical  condensers  is  electrically  identical  with 
the  two  given  parallel  conductors  A  and  B. 

In  a  medium  of  constant  permittivity  the  resultant  stress  or 
voltage  gradient,  produced  at  a  point  by  the  combined  action  of 
two  or  more  independent  electric  systems,  is  equal  to  the  geometric 
sum  of  the  stresses  produced  at  the  same  point  by  each  system. 


CHAP.  XVI]       ELASTANCE  OF  CABLES  AND  LINES 


177 


This  principle  of  superposition  can  be  considered  either  as  an  ex- 
perimental fact  or  as  an  immediate  consequence  of  the  fact  that 
in  a  medium  of  constant  permittivity  the  effects  are  proportional 
to  the  causes.  This  principle  being  true  for  electric  intensities, 
the  component  flux  densities  at  a  point  are  also  combined  accord- 
ing to  the  parallelogram  law,  because  they  are  proportional  to 
the  intensities.  Hence,  the  resultant  electrostatic  flux  can  be  re- 
garded as  the  result  of  the  superposition  of  the  fluxes  created  by 


FIG.  50. 


The  electrostatic  field  produced  by  a  single-phase  trans- 
mission line. 


the  component  systems.  Furthermore,  the  actual  voltage  be- 
tween any  two  points  in  the  dielectric  is  the  algebraic  sum  of  the 
voltages  due  to  the  component  systems,  because  each  voltage  is 
the  line  integral  of  the  corresponding  voltage  gradient,  and  the 
principle  of  superposition  is  valid  for  these  gradients.  This  line 
integral  is  a  function  only  of  the  positions  of  the  two  points,  and 
is  independent  of  the  path  along  which  the  integration  is  per- 
formed. This  latter  fact  is  very  convenient  in  applications  of  the 
principle  to  the  solution  of  problems. 


178  THE  ELECTRIC  CIRCUIT  [ART.  60 

In  order  to  be  able  to  apply  the  formulae  deduced  in  the  pre- 
ceding article,  it  is  essential  that  the  diameters  of  the  two  wires  be 
small  as  compared  with  the  distance  between  them.  The  reason  is 
that  each  component  system  is  supposed  to  possess  a  radial  field, 
in  spite  of  the  presence  of  the  other  conductor.  This  is  practically 
true  when  the  second  conductor  is  so  small,  or  so  far  distant  from 
the  first,  that  the  infinite  permittivity  of  its  material  does  not 
appreciably  distort  the  radial  field.  To  be  more  precise,  the  dis- 
tortion of  the  radial  component  field  originating  from  each  con- 
ductor, caused  by  the  presence  of  the  other,  must  be  negligible. 

It  is  sufficient  to  calculate  the  elastance  of  that  part  of  the 
system  between  one  of  the  conductors  and  the  neutral  plane  of 
symmetry  00',  the  total  elastance  being  equal  to  twice  that 
value.  This  we  can  do  by  computing  the  voltage  needed  to 
produce  a  displacement  Q'  per  unit  length  of  the  line.  The  volt- 
age between  the  surface  of  the  conductor  A  and  the  point  N  in 
the  plane  of  symmetry  00'  is  equal  to  \  E.  On  the  other  hand, 
the  .same  voltage  can  be  expressed  as  the  sum  of  the  voltages  due 
to  the  two  component  systems.  Referring  to  eq.  (290),  let  the 
distance  r'  refer  to  the  point  N,  and  let  r  refer  to  a  point  on  the 
surface  of  the  conductor  A.  Then,  as  far  as  the  first  component 
system  is  concerned,  the  voltage  between  A  and  N  is  equal  to 
(aQ'/lv}  Ln  (\  b/a),  where  Qf  is  the  actual  displacement  per  unit 
length  of  the  line.  In  the  second  system,  the  voltage  between  the 
same  two  points  is  —  (ffQ'/2v)  Ln  (\b/b}.  The  minus  sign  is 
due  to  the  fact  that  the  displacement  in  the  second  system  is 
toward  the  conductor  B,  and  hence  must  be  considered  as  negative 
if  that  at  the  first  conductor  is  regarded  as  positive.  The  ratio 
r' IT  for  the  second  system  is  more  accurately  equal  to  \  b/(b  —  a), 
but  a,  being  by  supposition  small  as  compared  to  6,  is  neglected  in 
the  denominator.  Equating  the  sum  of  the  preceding  two  expres- 
sions for  the  voltage  between  A  and  N  to  the  actual  voltage  £  E, 
we  get 

\E  =  (aQ'/2  TT)  Ln  (b/a) (295) 

Hence,  the  elastance  between  one  of  the  conductors  and  the 
neutral  plane  00',  for  a  unit  of  axial  length,  is 

S1  =  (C")-1  =  \  E/Q'  =  (a/2  T)  Ln(6/o),       .     (296) 
or,  with  air  as  the  dielectric, 

S'  =  (C')-1  =  41.45  log  (6/a)  megadarafs  per  kilometer.     (297) 


CHAP.  XVI]       ELASTANCE  OF  CABLES  AND  LINES  179 

The  corresponding  permittance  is 
C'  =  (S')~l  =  0.0241/log  (b/a)  microfarads  per  kilometer.    (298) 

When  using  these  formulae,  one  must  not  forget  that  the  per- 
mittance is  proportional  to  the  length  of  the  line,  while  the  elas- 
tance  varies  inversely  as  the  length  of  the  line.  The  total  elastance 
for  a  unit  length  between  the  two  conductors  is  equal  to  2  Sf,  the 
corresponding  permittance  being  \  C'. 

Prob.  1.  For  a  few  standard  spacings  and  sizes  of  conductor,  check 
the  values  of  permittance  given  by  eq.  (298)  with  those  tabulated  in  an 
electrical  pocketbook. 

Prob.  2.  For  some  assumed  values  of  a,  6,  and  E,  corresponding  to 
an  actual  transmission  line,  plot  a  curve  of  values  of  the  voltage  gradient 
along  the  line  A  B,  and  also  draw  the  horizontal  straight  line  represent- 
ing the  average  gradient  E/b. 

Hint :  At  a  distance  x  from  A  the  intensity  due  to  the  system  A  is 
aQ'/  (2  TTZ);  that  due  to  the  system  B  is  oQ' /  [2  *•  (6  -  a;)],  both  inten- 
sities being  directed  from  left  to  right. 

Prob.  3.  In  Fig.  50  let  A  and  B  be  small  spheres,  instead  of  cylinders. 
Show  that  the  elastance  between  one  of  the  spheres  and  the  neutral 
plane  00'  is  equal  to  (<r/4  *•)  (I/a  —  1/6). .  Hint:  Apply  the  principle  of 
superposition,  as  in  the  text  above,  and  utilize  the  solutions  of  problems 
11  and  12  of  the  preceding  article. 

Prob.  4.  In  a  transmission  line  the  wire  B  is  split  into  two  separate 
conductors  BI  and  B2,  connected  in  parallel.  The  spacings  A  —  BI, 
A  —  B2,  and  BI  —  52  are  equal  to  6t,  62,  and  612  respectively.  Show  how 
to  calculate  the  total  permittance  per  unit  length  of  the  line,  using  the 
method  of  superposition.  Solution:  Let  Qf,  Qi,  and  Q2'  be  the  displace- 
ments issuing  per  unit  length  of  the  conductors  A,  BI,  and  B2  respectively. 
Resolve  the  given  system  into  three  systems,  with  the  three  given  con- 
ductors each  concentric  with  a  cylinder  of  infinite  radius.  Then  we  have 
the  following  three  conditions :  (a)  Qf  +  Q/  +  Q2'  =  0,  because  electricity 
behaves  as  an  incompressible  fluid;  (b)  the  given  voltage  E  between 
A  and  BI  is  the  sum  of  the  partial  voltages  for  the  three  component 
systems,  each  expressed  according  to  eq.  (290) ;  and  (c)  the  same  is  true 
for  the  voltage  E  between  A  and  Bz.1  From  these  three  equations  the 
quantities  Qi  and  Q2'  are  eliminated,  and  the  required  elastance  is  de- 
termined from  the  resultant  equation,  as  the  ratio  of  E  to  Q'. 

Prob.  6.  Show  how  to  calculate  the  elastance  between  two  small 
cylinders  or  spheres  of  unequal  radii. 

Prob.  6.  Analyze  the  formal  mathematical  reason  for  which  the  elec- 
trostatic equipotential  lines  in  Figs.  49  and  50  coincide  with  the  magnetic 
lines  of  force,  and  vice  versa.  Compare  Figs.  46  and  47,  Arts.  59  and 
60,  in  the  Magnetic  Circuit. 

1  Or  else  we  may  use  as  condition  (c)  the  fact  that  the  resultant  voltage 
between  BI  and  J52  equals  zero. 


180  THE  ELECTRIC  CIRCUIT  [ART.  61 

61.1  The  Influence  of  the  Ground  upon  the  Elastance  of  a 
Single-phase  Line.  When  the  ground  is  used  as  the  return  con- 
ductor of  a  circuit,  for  instance  in  single-phase  railways  and  in 
telegraph  lines,  the  elastance  of  the  circuit  is  calculated  by  assum- 
ing the  ground  to  be  a  good  conductor  of  electricity;  in  other 
words,  its  permittivity  is  assumed  to  be  infinitely  large.  This 
gives  a  larger  permittance  than  any  other  assumption,  and  con- 
sequently a  value  which  is  on  the  safe  side.  According  to  the 
law  of  refraction  (Art.  55)  the  lines  of  force  from  the  metallic  con- 
ductor enter  the  ground  at  right  angles  to  its  surface;  so  that  the 
field  has  the  shape  shown  in  Fig.  50,  between  one  of  the  wires 
and  the  plane  of  symmetry  00',  which  in  this  case  represents  the 
surface  of  the  ground.  This  leads  to  Lord  Kelvin's  method  of  elec- 
tric images,  which  we  shall  use  in  its  simplest  form  only. 

When  it  is  required  to  find  the  shape  of  the  field,  or  the  elastance 
between  a  conductor  such  as  A  and  an  infinite  conducting  surface 
such  as  00' ,  first  locate  a  fictitious  conductor  B,  which  is  the  elec- 
tric image  of  the  conductor  A ;  that  is,  B  is  located  as  if  it  were 
the  optical  image  of  A  in  the  plane  mirror  00'.  Furthermore,  if 
A  has  a  potential  of  E  volts  above  that  of  00',  take  the  potential 
of  B  as  E  volts  below  that  of  00',  the  voltage  between  A  and  B 
thus  being  2  E  volts.  Having  located  B,  the  conducting  plane 
00'  is  removed,  and  the  field  between  A  and  B  is  determined. 
The  part  of  the  field  between  A  and  00'  has  real  existence,  that 
between  00'  and  B  is  fictitious. 

The  validity  of  this  principle  in  the  case  under  consideration 
becomes  evident  by  the  following  reasoning:  Let  a  voltage  of 
2  E  be  maintained  between  A  and  B  by  means  of  a  battery. 
Place  an  infinite  conducting  sheet  of  negligible  thickness  so  as 
to  coincide  with  the  equipotential  plane  00'.  The  field  is  not 
affected  thereby,  the  lines  of  displacement  being  normal  to  this 
sheet.  Connecting  the  sheet  to  the  middle  point  of  the  battery 
does  not  in  any  way  disturb  the  field.  Now  the  field  between  A 
and  the  sheet  00'  is  maintained  by  one  half  of  the  battery,  that 
between  B  and  00'  by  the  other  half.  Both  halves  are  in  equi- 
librium independently  of  each  other,  so  that  the  conductor  B  with 
its  half  of  the  battery  may  be  removed  without  disturbing  the 
field  between  A  and  00'.  Conversely,  to  find  the  field  between 

1  The  rest  of  this  chapter  may  be  omitted  if  so  desired,  as  it  is  not  neces- 
sary to  an  understanding  of  the  remainder  of  the  book. 


CHAP.  XVI]       ELASTANCE  OF  CABLES  AND  LINES  181 

A  and  a  conducting  sheet  00',  the  latter  is  replaced  by  a  fictitious 
conductor  B,  so  as  to  reduce  the  conditions  to  those  investigated 
in  the  previous  article.  For  a  general  discussion  of  the  principle 
of  electric  images,  see  any  standard  work  on  the  mathematical 
theory  of  electricity  and  magnetism. 

Applying  this  principle  to  the  case  of  a  single-phase  line  with 
ground  return,  we  see  immediately  that  all  of  the  formulae  de- 
duced in  the  preceding  article  hold  true,  provided  that  we  put 
6  =  2h,  where  h  is  the  elevation  of  the  conductor  above  the 
ground.  Since  this  elevation  is  usually  quite  considerable,  it  will 
be  seen  that  the  elastance  of  the  circuit  is  larger,  and  the  charging 
current  smaller  as  compared  to  the  case  of  a  metallic  circuit 
having  a  comparatively  small  spacing. 

The  next  case  to  be  considered  is  that  of  the  elastance  of  a 
metallic  return  line,  as  reduced  by  the  proximity  of  the  earth 
(Fig.  51).  The  elastance  is  reduced  as  compared  to  that  in  Fig. 
50  because  part  of  the  medium  of  finite  elastivity  (air)  is  replaced 
by  the  ground,  which  is  assumed  to  be  of  zero  elastivity,  or  a 
good  conductor  of  electricity.  It  will  be  seen  from  the  figure  that 
the  lines  of  force  are  deflected  toward  the  ground,  where  they 
find  a  path  of  less  elastance. 

The  total  elastance  between  the  conductors  A  and  B  is  cal- 
culated, using  again  the  method  of  electric  images.  The  lines  of 
force  meet  the  ground  at  right  angles;  and  its  surface  is  one  of 
equal  potential.  The  field  above  the  ground  would  be  the  same 
if  the  ground  were  removed  and  replaced  by  the  electric  images 
A'  and  B'  of  the  wires,  the  polarity  of  the  images  being  that  indi- 
cated in  the  sketch.  The  surface  of  the  ground  becomes  now 
a  plane  of  symmetry.  The  fictitious  field  below  the  ground  is 
indicated  by  the  dotted  lines.  The  dielectric  field  may  now  be 
considered  as  if  due  to  a  superposition  of  four  systems,  each  consist- 
ing of  one  of  the  conductors  and  a  cylinder  at  infinity.  Applying 
eq.  (290)  for  each  of  these  four  systems,  we  find  that  the  voltage 
between  A  and  B, 

due  to  system  A,  is  +(<rQ'/2  T)  Ln  (b/a); 
due  to  system  B,  is  —  (vQ'/2  TT)  Ln  (a/6) ; 
due  to  system  A',  is  -  (<rQ'/2  TT)  Ln  (2  d/2  hi} ; 
due  to  system  Br,  is  +(oQ'/2  TT)  Ln  (2  h2/2  d) ; 

where  2  d  =  AB'  =  A'B.     The  actual  voltage  between  A  and  B 


1S2 


THE  ELECTRIC  CIRCUIT 


[ART.  61 


being  equal  to  E,  we  have,  by  adding  the  preceding  four  ex- 
pressions, 

E  =  (<rQ'/2  TT)  [2  Ln  (6/a)  -  Ln  (tf/hJk)].  .     .     (299) 

One  half  of  the  elastance  between  A  and  B  is 

Sr  =  i-  E/Q'  =  (cr/27r)  [Ln  (b/a)  -  %  Ln  (d2/W].     (300) 


FIG.  51.     The  electrostatic  field  due  to  a  single-phase  line  AB,  as  affected 
by  the  proximity  of  the  ground.    A'  and  B'  are  the  electric  images  of  A  and  B. 

This  expression  is  identical  with  formula  (296),  except  for  the  last 
term,  which  represents  the  reduction  in  elastance  due  to  the  pres- 
ence of  the  ground.  When  the  distances  h  and  k*  to  the  ground  are 
large  as  compared  to  the  spacing  6,  the  ratio  tf/fuhe  differs  but 
ittle  from  unity,  and  the  correction  is  small  because  the  logarithm 


CHAP.  XVI]       ELASTANCE  OF  CABLES  AND  LINES  183 

of  unity  is  equal  to  zero.  Equation  (300)  can  be  written  also  in 
a  simpler  form  by  combining  the  two  logarithms  into  one.  We 
obtain  then 

S'  =  (<r/2x)Ln(6c/a), (301) 

where  bc  stands  for  the  corrected  spacing,  determined  from  the  ex- 
pression 

bc  =  b(Vhihv/d).   .    .    . .  .    .    .     (302) 

We  have  thus  arrived  at  the  following  simple  rule:  The  elastance 
and  permittance  of  a  single-phase  line,  with  the  effect  of  the  ground 
considered,  are  expressed  by  the  same  formulae  (296)  to  (298)  as 
though  this  effect  were  ignored,  provided  that  the  actual  spacing  b  is 
replaced  by  the  corrected  spacing  bc  given  by  formula  (302)  or  (305). 
In  practice,  the  values  of  hi,  h%,  and  6  are  known,  and  it  is  de- 
sirable to  avoid  the  use  of  the  quantity  d  in  the  foregoing  formula. 
Applying  a  well-known  theorem  of  elementary  geometry,  we  have 
from  the  triangle  AA'B 

A7!!2  =  lA'2  +  AB2  +  2  AA'  X  IF, 
or 

4d2  =  4  ^2  +  62  +  4Ai(fo-fo), 

from  which 

4  dz  =  b2  +  4  hjtt. 
Hence 

d*/hih  =  1  +  i  V/hih* (303) 

Equation  (300)  becomes  then 

S'  =  (a/2  T)  [Ln  (6/a)  -  J  Ln  (1  +  J  V/hJ*)],  .     (304) 
and  from  eq.  (302)  we  have 

bc  =  b/Vl  +  i  b*/hiht. (305) 

When  tables  of  capacity  for  standard  spacings  are  used,  as  tabu- 
lated in  various  reference  books,  the  correction  for  the  influence 
of  the  ground  will  be  found  convenient  in  the  form  shown  in 
problem  2  below. 

Prob.  1.  For  various  usual  spacings  and  sizes  of  conductor,  calculate 
the  per  cent  error  introduced  in  computing  the  permittance  of  a  trans- 
mission line  by  neglecting  the  influence  of  the  earth  in  the  most  unfavor- 
able cases.  Select  the  conductors  either  in  a  vertical  or  in  a  horizontal 
plane,  whichever  arrangement  in  your  opinion  is  more  affected  by  the 
ground. 


184  THE  ELECTRIC  CIRCUIT  [ART.  62 

Prob.  2.  When  permittances  are  taken  from  standard  tables,  it  is  not 
convenient  to  use  the  corrected  spacing  bc,  because  capacities  are  tabu- 
lated for  standard  spacings  only.  In  this  case  it  is  convenient  to  repre- 
sent the  elastance  given  by  eq.  (304)  in  the  form  Sc'  =  S'  -  s,  where  S' 
is  the  reciprocal  of  the  value  of  capacity  found  in  the  tables,  and  s  is  the 
correction  due  to  the  presence  of  the  ground.  Deduce  a  simple  form  of 
this  correction,  when  it  is  small.  Solution:  Expanding  the  natural  loga- 
rithm according  to  the  series  Ln  (1  +  x)  =  x  —  f  x2  +  i  z3  —  etc.,  we  find 
that  the  correction  s  =  9  Ln  (1  +  \  fr/h^)  =  9  [|  fr/h^  -  |(i  &VW2 
+  etc.]  in  megadarafs  per  kilometer  of  one  conductor.  S'  must  of  course 
be  taken  also  in  megadarafs  per  kilometer  of  one  conductor. 

Prob.  3.  Deduce  formulae  for  the  influence  of  the  ground  in  the  case 
of  small  spheres  in  place  of  the  cylinders. 

62.  The  Equations  of  the  Electrostatic  Lines  of  Force  and 
Equipotential  Surfaces  Produced  by  a  Single-phase  Line.  In 

Fig.  50,  let  P  be  a  point  on  the  line  of  force  AN'PB,  of  which 
we  desire  to  find  the  equation.  Let  us  calculate  the  total  flux 
which  passes  from  conductor  A  to  B  between  the  plane  of  sym- 
metry A  B,  and  the  surface  of  force  on  which  the  point  P  is 
located.1 

Let  the  axial  length  for  which  the  flux  is  determined  be  equal 
to  one  centimeter.  This  flux  may  be  considered  as  the  resultant 
of  the  fluxes  due  to  the  systems  A  and  B.  The  radial  flux  pass- 
ing between  AB  and  P,  due  to  the  component  system  having 
the  center  A,  is  equal  to  Q'6i/2ir,  and  is  directed  from  left  to 
right.  The  flux  due  to  the  B  system  is  equal  to  Q'dz/2  IT,  and  is 
also  directed  from  left  to  right,  B  being  the  negative  conductor. 
The  total  or  the  actual  flux  between  the  surfaces  of  force  AB  and 
AN'PB  equals  (Q'/2  ir)  (0i  +  02).  Since  this  flux  does  not  depend 
upon  the  position  of  the  point  P,  provided  that  the  point  is  taken 
upon  the  line  of  force  under  consideration,  we  have 

0l  -f  02  =  const (306) 

for  all  points  on  a  line  of  force.  This  is  the  equation  of  the  line 
of  force.  For  different  lines  of  force,  the  value  of  the  constant  is 
different.  In  the  triangle  APB,  the  angle  w  is  supplementary  to 

1  It  is  convenient  to  speak  of  a  surface  formed  by  lines  of  force  as  a  sur- 
face of  force.  For  instance,  a  line  of  force  such  as  AN'PB,  should  it  move  in 
a  direction  perpendicular  to  the  plane  of  the  paper,  would  form  a  cylindrical 
surface,  which  we  shall  call  a  surface  of  force,  by  analogy  with  a  line  of  force. 
On  the  other  hand,  we  shall  call  a  line  such  as  CPC'  an  equipotential  line,  to 
distinguish  it  from  the  corresponding  equipotential  surface  which  it  repre- 
sents in  the  sketch. 


CHAP.  XVI]       ELASTANCE  OF  CABLES  AND  LINES  185 

the  sum  of  the  angles  61  and  02,  so  that  condition  (306)  may  also 
be  written 

w  =  const (307) 

This  represents  the  arc  of  a  circle  passing  through  A  and  B,  of 
which  w  is  the  inscribed  angle.  It  is  thus  proved  that  the  lines  of 
force  are  arcs  of  circles  passing  through  A  and  B. 

For  points  on  the  line  of  symmetry  00'  the  angles  0i  and  02  are 
equal,  so  that  the  total  flux  corresponding  to  a  certain  angle  0i 
is  (Q'/2  TT)  (2  0i)  =  Q'0i/7r.  This  fact  permits  us  to  mark  on  the 
line  00'  the  intersections  of  the  surfaces  of  force  between  which 
are  included  definite  fractions  of  the  total  flux  Q'.  For  instance, 
let  it  be  desired  to  draw  a  line  of  force  such  that  the  flux  between 
it  and  the  plane  A  B  shall  equal  one  sixth  of  the  total  flux.  One 
sixth  of  180  degrees  are  30  degrees;  we  therefore  draw  from  A  a 
straight  line  at  an  angle  of  30  degrees  to  AB,  and  through  its  inter- 
section with  00'  draw  an  arc  of  a  circle  passing  through  A  and  B. 
In  this  way,  the  total  flux,  or  what  is  the  same,  the  total  per- 
mittance between  A  and  B,  can  be  divided  into  any  number  of 
equal  or  unequal  permittances  in  parallel. 

To  prove  that  the  equipotential  lines  are  also  circles,  take 
again  a  point  P  determined  by  the  distances  n  and  r2  from  A  and 
B  respectively.  If  the  point  C  lies  on  the  same  equipotential  line, 
the  voltage  between  P  and  C  is  equal  to  zero,  so  that,  applying 
eq.  (290)  for  the  two  component  systems,  we  get 

(aQ72  TT)  Ln  (n/AC)  -  (*Q'/2  *•)  Ln  (r2/£C)  =  0, 

from  which 

Ln  (n/AC)  =  Ln  (r2/BC), 
or 

ri/r2  =  AC/BC  =  const (308) 

This  is  the  equation  of  an  equipotential  line  in  "  bipolar  "  co- 
ordinates; the  curve  is  such  that  the  ratio  of  TI  to  r2  remains 
constant.  This  constant  is  different  for  each  equipotential  line, 
because  each  line  has  its  own  point  C. 

Equation  (308)  may  be  proved  to  represent  a  circle,  by  select- 
ing an  origin,  say  at  A,  and  substituting  for  r\  and  rz  their  values 
in  terms  of  the  rectangular  coordinates  x  and  y.  The  following 
proof  by  elementary  geometry  leads  to  the  same  result.  Produce 
AP  and  lay  off  PD  =  PB  =  rz.  According  to  eq.  (308),  BD  is 
parallel  to  CP,  and  consequently  PC  bisects  the  angle  APB  —  <a. 


186  THE  ELECTRIC  CIRCUIT  [ART.  62 

Let  the  point  C'  lie  on  the  same  equipotential  line  with  C;  then 
the  voltage  between  P  and  C'  is  also  equal  to  zero,  and  by  analogy 
with  eq.  (308)  we  have 

ri/r2  =  AC'/BC'  =  const  .....     (309) 

By  plotting  PD'  =  r2  (not  shown  in  the  figure)  along  PA,  in  the 
opposite  direction  from  PD,  and  connecting  D'  to  B,  one  can  show 
as  before  that  PC'  bisects  the  angle  BPD  =  180°  -  co.  But  the 
bisectors  of  two  supplementary  angles  are  perpendicular  to  each 
other;  consequently,  CPC'  is  a  right  angle,  and  the  point  P  lies 
on  a  semicircle  drawn  on  the  diameter  CC'.  This  semicircle  is 
the  equipotential  line  itself,  because  all  points,  such  as  P,  which 
are  determined  by  C  and  C",  must  lie  on  it.  The  semicircle  below 
AB  evidently  belongs  to  the  same  equipotential  line. 

From  eqs.  (308)  and  (309)  the  following  expressions  are  ob- 
tained for  the  radius  R  of  the  equipotential  line  under  considera- 
tion. 

RC 

R  =  i-(Bc/Acy         •  •  •  <310> 

or 


1  +  BC'/AC','      ..... 

so  that  the  equipotential  line  can  be  easily  drawn  for  a  given 
C  or  C'. 

Let  it  be  required  to  calculate  the  elastance  of  the  slice  of  die- 
lectric between  the  neutral  plane  00'  and  the  equipotential  sur- 
face passing  through  a  given  point  C.  It  is  sufficient  to  find  the 
expression  for  a  unit  axial  length,  knowing  that  the  elastance  is 
inversely  proportional  to  the  length  of  the  conductors. 

Write  the  expression  for  the  voltage  between  the  points  N  and 
C,  using  again  eq/  (290).  For  the  systems  A  and  B  we  have 

ENC  =  (*Q'/2ir)  [Ln  (AC  /AN)  -  Ln  (BC/BN}}, 
or,  since  AN  =  BN, 

ENC  =  (<rQ'/2ir)Ln(AC/BC)  .....     (312) 
From  this  equation  we  see  that  the  elastance  per  centimeter 

Sffc'.=  ENC/Q'  =  (a/2  TT)  Ln  (AC/BC}.     .     .     (313) 

From  this  expression,  the  elastance  between  any  two  equipotential 
surfaces  can  be  calculated,  by  computing  first  the  elastance  be- 


CHAP.  XVI]       ELASTANCE  OF  CABLES  AND  LINES  187 

tween  each  of  the  surfaces  and  the  plane  of  symmetry  00',  and 
then  taking  either  the  sum  or  the  difference  of  these  elastances, 
depending  upon  the  positions  of  the  two  given  surfaces;  that  is, 
whether  they  lie  on  different  sides  or  on  the  same  side  of  the 
plane  00'. 

It  has  been  explained  above  how  to  divide  the  field  by  surfaces 
of  force  into  permittances  of  desired  values,  these  permittances  be- 
ing proportional  to  the  angles  0i  or  02,  determined  by  the  point  Nf 
on  the  neutral  plane.  Knowing  now  how  to  subdivide  the  field 
into  elastances  of  desired  values  by  equipotential  surfaces,  the 
student  can  without  difficulty  calculate  the  permittance  or  the 
elastance  of  a  given  slice  in  the  field  between  two  equipotential 
surfaces  and  two  surfaces  of  force. 

Prob.  1.  For  an  assumed  size  of  the  conductors  and  a  spacing  used  in 
extra-high-tension  transmission  lines,  draw  a  set  of  lines  of  force  and  equi- 
potential lines  (Fig.  50)  such  as  to  divide  the  total  voltage  and  the  total 
electrostatic  flux  into  10  equal  parts. 

Prob.  2.  Let  A  and  B  (Fig.  50)  be  two  very  small  wires  at  a  distance 
of  90  cm.  between  centers.  What  is  the  permittance  of  the  slice  NN'PC 
if  ffN'  =  25  cm.;  NC  =  32  cm.;  and  the  axial  length  is  180  m.? 

Ans.    0.000861  mf. 

Prob.  3.  Show  that  the  lines  of  force  between  two  small  spheres  are 
not  circles,  but  curves  the  equation  of  which  is  cos  0t  +  cos  02  =  const. 

Prob.  4.  Show  that  the  equipotential  surfaces  in  the  case  of  two 
small  spheres  are  represented  by  the  equation  1/ri  —  l/r2  =  const. 

Prob.  5.  Show  how  to  draw  in  a  given  case  the  field  shown  in  Fig.  51. 
Solution:  Draw  a  set  of  n  lines  of  force  due  to  the  system  AB  alone,  the 
same  as  in  Fig.  50.  Let  the  flux  Q'  be  divided  into  n  equal  parts,  so  that 
the  flux  between  the  adjacent  surfaces  of  force  is  Q'/n.  Draw  a  similar 
set  of  lines  of  force  for  the  system  A'B'.  The  equation  of  a  line  of  force 
in  the  system  AB  is  w  =  const.;  that  in  the  system  A'B'  is  u>  =  const. 
According  to  the  principle  of  superposition,  the  equation  of  a  line  of  force 
in  the  resultant  field  is  w  —  w'  =  const.,  the  minus  sign  being  due  to  the 
fact  that  A'  is  negative  if  A  is  positive.'  Let  the  point  of  intersection  of 
two  lines  of  force  w  =  C  and  «'  =  C"  be  the  starting,  point  for  draw- 
ing a  line  of  force  in  the  resultant  field.  Then  the  point  of  intersection 
of  the  next  lines  «  =  C  +  w/n  and  J  =  C'  +  v/n  also  belongs  to  the 
same  line  of  force  in  the  resultant  field,  because  for  both  points  u  —  u'  = 
C  —  C'.  In  other  words,  the  lines  of  force  in  the  resultant  field  are 
diagonal  curves  with  respect  to  the  lines  of  force  in  the  component  fields, 
and  may  be  drawn  from  intersection  to  intersection.  A  similar  construc- 
tion holds  for  equipotential  surfaces.  The  student  is  strongly  urged  to 
try  this  construction  for  some  assumed  data,  because  the  method  of 
diagonal  curves  is  generally  applicable  when  a  given  field  can  be  resolved 
into  two  simpler  fields. 


188  THE  ELECTRIC  CIRCUIT  [ART.  63 

63.  The  Elastance  between  Two  Large  Parallel  Circular 
Cylinders.  The  formulae  derived  in  Art.  60,  for  the  elastance 
and  permittance  of  a  homogeneous  medium  between  two  parallel 
cylinders,  hold  true  only  when  the  diameters  of  the  cylinders  are 
small  as  compared  to  the  interaxial  distance,  for  the  reason  there 
explained.  When  the  diameters  of  the  "cylinders  are  compara- 
tively large,  the  elastance  is  derived  by  reducing  the  conditions  to 
those  obtaining  in  Art.  60. 

Let  A  and  B  (Fig.  50)  represent  as  before  two  conductors  of 
very  small  diameter,  and  let  a  difference  of  potential  of  100  volts 
be  maintained  between  them  by  means  of  a  battery.  Let  the  volt- 
age between  the  conductor  B  and  the  equipotential  surface  CPC' 
be  20  volts.  Place  an  infinitely  thin  metal  sheet  so  as  to  coincide 
with  this  surface,  and  connect  this  sheet  to  a  point  of  the  battery 
such  that  the  voltage  between  it  and  the  conductor  B  still  remains 
equal  to  20  volts.  These  changes  do  not  affect  the  electrostatic 
field  either  inside  or  outside  the  surface  CPC',  the  displacement 
being  normal  to  this  surface.  Now  remove  the  conductor  B  al- 
together, leaving  a  difference  of  potential  of  80  volts  maintained 
by  the  battery  between  the  conductor  A  and  the  cylinder  CPC'. 
The  field  outside  the  cylinder  is  not  affected;  that  inside  of  it  has 
entirely  disappeared.  We  have  now  a  field  between  the  cylinder 
A  of  very  small  diameter  and  the  cylinder  CPC'  of  a  compara- 
tively large  diameter.  Take  now  another  equipotential  surface, 
for  instance  KMK',  symmetrical  with  CPC',  place  a  metal  cylinder 
so  as  to  coincide  with  it,  and  connect  it  to  a  tap  on  the  battery, 
so  that  the  same  difference  of  potential  of  20  volts  remains  be- 
tween this  cylinder  and  the  conductor  A.  The  field  is  not  altered 
by  this  connection,  and  now  the  conductor  A  may  be  removed. 
Thus,  we  finally  obtain  a  field  between  two  cylinders  of  compara- 
tively large  diameter.  The  difference  of  potential  between  the 
cylinders  is  only  60  volts,  while  the  original  difference  of  potential 
between  the  conductors  A  and  B  was  100  volts. 

Conversely,  let  the  cylinders  CPC'  and  KMK'  be  given,  and 
let  it  be  required  to  find  the  shape  of  the  field  between  them,  and 
the  elastance  of  this  field.  The  problem  is  reduced  to  that  of 
finding  the  positions  of  the  infinitely  small  eccentric  conductors 
A  and  B,  with  respect  to  which  the  given  cylinders  are  equi- 
potential surfaces.  Then  the  field  is  mapped  out  according  to 
the  formulae  given  in  the  preceding  article,  leaving  out  the  space 


CHAP.  XVI]       ELASTANCE  OF  CABLES  AND  LINES  189 

inside  the  cylinders.  The  elastance  between  one  of  the  large 
cylinders  and  the  plane  00'  is  calculated  by  using  formula  (313). 
This  method  is  applicable  whether  the  two  cylinders  are  of  the 
same  radius  or  not,  and  whether  one  is  outside  or  inside  of  the 
other.  It  is  always  possible  to  find  the  positions  of  the  lines  A 
and  B  with  respect  to  which  the  given  cylinders  represent  equi- 
potential  surfaces.  The  details  of  the  calculation  are  given  below. 
Consider  first  the  case  of  two  cylinders  CPC'  and  KMK'  of 
the  same  diameter  d;  let  the  distance  between  the  centers  p  and 
q  of  these  cylinders  be  equal  to  c.  In  order  to  use  eq.  (313),  it  is 
necessary  to  express  AC  and  BC  through  the  given  quantities  c 
and  d.  According  to  eqs.  (308)  and  (309),  we  have 

AC/BC  =  AC'/BCf (314) 

All  the  quantities  which  enter  into  this  equation  can  be  expressed 
through  one  unknown  length,  for  instance  BC.  We  put 

BC  =  AK  =  x] 
then 

AC  =  CK  +  AK  =  (c  -  d)  +  X-, 
AC'  =  x  +  c;  -     •     •     (315) 

BC'  =  d  -  x. 

Substituting  these  values  into  eq.  (314),  and  solving  the  resulting 
quadratic  equation  for  x,  we  obtain,  retaining  the  positive  value 
only,  

=  %d[-(a-l)  +  \/«2-  1], (316) 

where  the  ratio  of  the  interaxial  distance  to  the  diameter  is  de- 
noted by  a,  or 

a  =  c/d (317) 

By  substituting  this  value  of  x  into  the  expression  for  AC  m  eqs. 
(315),  we  find 


AC  =  i[(c  -  d)  +  Vc2  -  d2]  =  id  [(a  -  1)  +  W  -  1],     (318) 
so  that 


AC/BC  =  [(a  -  1)  +  V«2  -  !]/[  _  («  -  1)  4-  Va2  -  1]. 

This  expression  can  be  simplified  by  multiplying  both  the  numer- 
ator and  the  denominator  by  the  value  of  the  numerator,  so 


190  THE  ELECTRIC   CIRCUIT  [ART.  63 

as  to  get  rid  of  the  square  root  in   the   denominator.      The 

result  is  

AC/BC  =  a  +  V«2  -  1 (319) 

The  expression  (313)  for  the  elastance  between  one  of  the  cylinders 
and  the  plane  of  symmetry,  per  unit  of  axial  length,  becomes 

S'  =  (0/2*)  Ln  [a  +  Vo2^!].       .     .     .     (320) 

Those  familiar  with  hyperbolic  functions  will  notice  that  the  pre- 
ceding equation  can  be  simplified  into 

S'  =  (cr/271-)  Cosh-1** (321) 

Since  tables  of  hyperbolic  functions  are  readily  available,  the 
evaluation  of  elastance  is  simpler  in  this  form  than  it  is  if  eq.  (320) 
is  used.1 

When  the  diameter  of  the  conductors  is  small  as  compared  to 
the  interaxial  distance,  a  is  a  large  quantity,  and  unity  under  the 
radical  sign  in  eq.  (320)  may  be  neglected.  This  equation  be- 
comes then  practically  identical  with  eq.  (296).  For  large  values 
of  a,  the  term  (1  —  I/a2)*,  obtained  by  factoring  in  expression  (320), 
is  conveniently  expanded  according  to  the  binomial  theorem,  the 
result  being 

S'  =  (a/2 TT)  Ln  (2  a  -  i  a"1  -  |  or3  -  TV  a~5 -   .  .  .).     (322) 

With  the  exception  of  2  a,  all  of  the  terms  in  parentheses  are 
small  corrections  to  the  result. 

Let  now  the  diameters  of  the  two  given  cylinders  be  different. 
In  addition  to  relation  (314),  we  also  have 

BK/AK  =  BK'/AK' (323) 

It  is  necessary  in  this  case  to  introduce  two  unknown  quantities, 
BC  =  x  and  AK  =  y.  Equations  (315)  are  modified  accordingly. 
All  of  the  quantities  in  eqs.  (314)  and  (323)  are  expressed  through 
x  and  y,  and  then  these  two  equations  are  solved  together  for  x 
and  y.  After  this,  the  elastance  between  each  cylinder  and  the 
plane  00'  is  expressed  by  using  eq.  (313). 

1  Dr.  A.  E.  Kennelly,  "  The  Linear  Resistance  between  Parallel  Conduct- 
ing Cylinders  in  a  Medium  of  Uniform  Conductivity,"  Proceedings  Amer. 
Philosophical  Soc.,  Vol.  48  (1909),  p.  142;  also  his  article  on  "  Graphic  Rep- 
resentations of  the  Linear  Electrostatic  Capacity  between  Equal  Parallel 
Wires,"  Electrical  World,  Vol.  56  (1910),  p.  1000.  See  also  his  book  on  Ap- 
plications of  Hyperbolic  Functions  to  Electrical  Engineering  (1912). 


CHAP.  XVI]       ELASTANCE  OF  CABLES  AND  LINES  191 

In  some  cases  it  is  required  to  calculate  the  dielectric  flux 
density  at  a  point  in  the  field  between  the  cylinders,  or  at  the  sur- 
face of  one  of  the  cylinders.  Let  P  (Fig.  52)  be  a  point  in  the 
field  between  two  parallel  cylinders,  small  or  large  ;  the  flux  density 
at  P  is  the  geometric  sum  of  the  densities  due  to  the  systems  A 
and  B.  The  flux  density  due  to  the  system  A  is 


while  that  due  to  B  is 


These  component  densities  are  directed  as  shown  in  Fig.  52. 
The  resultant  density  D  is  directed  along  the  tangent  to  the  line 


b 

FIG.  52.    Dielectric  flux  density  at  a  point,  determined  by  the  method 
of  superposition. 

of  force  through  P.     From  the  preceding  two  equations,  we  have 
the  relation 


so  that  the  triangles  APB  and  Pmn  are  similar.  The  correspond- 
ing sides  are  marked  with  one,  two,  and  three  short  lines  respec- 
tively. From  these  triangles  we  can  write 

D  :  D,  =  b  :  r2, 
or,  substituting  the  foregoing  expression  for  DI, 

D  =  Q'6/(2W2)  .......     (324) 

From  this  expression,  the  flux  density  can  be  calculated  at  any 
point  in  the  field  or  on  the  surface  of  one  of  the  cylinders.  Mul- 
tiplying the  flux  density  by  the  elastivity  of  the  medium,  the  cor- 
responding dielectric  stress  is  obtained.  It  must  be  kept  well  in 
mind  that  b,  ri}  and  r2  refer  to  the  points  A  and  B,  and  not  to  the 
centers  p  and  q  of  the  actual  cylinders. 


192  THE  ELECTRIC   CIRCUIT  [ART.  63 

Prob.  1.  Take  two  equal  cylinders  at  a  comparatively  short  distance 
apart,  and  (a)  calculate  the  permittance  per  meter  of  the  axial  length; 
(b)  divide  the  field  into  10  equal  elastances  in  series  and  into  10  equal 
permittances  in  parallel;  (c)  plot  a  curve  of  the  flux  density  distribu- 
tion on  the  surface  of  one  of  the  cylinders. 

Prob.  2.  Show  that  on  an  equipotential  surface  surrounding  A,  and 
consequently  on  the  corresponding  metal  surface,  the  flux  density  varies 
inversely  as  r,2. 

Prob.  3.  Show  how  to  calculate  the  permittance  between  a  large  cylin- 
der and  a  given  infinite  plane. 

Prob.  4.  Show  that  A  and  B  are  inverse  points  with  respect  to  any 
equipotential  circle;  this  means  that  the  radius  qC  is  the  geometric  mean 
between  the  distances  qB  and  qA,  and  the  radius  pK  is  the  geometric 
mean  between  the  distances  pA  and  pB.  This  is  true  whether  the  radii 
qC  and  pK  are  equal  or  not. 

Prob.  5.  Extend  the  theory  given  in  this  article  to  the  calculation  of 
the  elastance  and  flux  density  distribution  between  two  large  spheres. 
Consult  the  chapters  on  electrostatics  in  some  standard  work  on  the 
mathematical  theory  of  electricity  and  magnetism. 


CHAPTER  XVII 

EQUIVALENT    ELASTANCE    AND  CHARGING  CURRENT 
OF  THREE-PHASE  LINES 

64.  Three-phase  Line  with  Symmetrical  Spacing.  Consider 
an  unloaded  three-phase  line,  and  let  the  three  conductors  be 
denoted  by  A,  B,  and  C  respectively.  There  is  a  displacement  of 
electricity  between  each  pair  of  conductors,  and  since  the  three  in- 
stantaneous voltages  are  different,  the  displacements  between  the 
three  pairs  of  conductors  at  any  instant  are  also  different.  The 
three  sets  of  lines  of  force  are  relatively  displaced  and  the  flux 
density  varies  from  instant  to  instant,  so  that  there  is  produced 
in  reality  a  revolving  electrostatic  field.  Let  the  instantaneous 
displacements  which  issue  from  the  three  conductors  per  unit  of 
axial  length  be  denoted  by  qi,  q2,  and  q3,  where  the  subscripts  1,  2, 
and  3  refer  to  the  conductors  A,  B,  and  C  respectively.  To  be 
consistent  with  the  notation  used  before,  these  symbols  should  be 
provided  with  the  "  prime  "  sign,  but  this  sign  is  omitted  in  order 
not  to  obscure  the  formulae.  The  displacements  are  considered 
positive  when  they  are  directed  from  the  conductors  into  the  die- 
lectric. Since  electricity  behaves  like  an  incompressible  fluid,  as 
much  of  it  as  is  displaced  at  any  instant  out  of  one  conductor 
must  be  displaced  into  the  other  two  conductors,  so  that  at  all 
times  the  following  relation  holds,  namely, 

ffi  +  ?2  +  ?3  =  0 (325) 

The  three  q's  vary  with  the  time  according  to  the  sine  law.  With 
a  symmetrical  spacing  of  the  wires,  and  symmetrical  voltages 
forming  an  equilateral  triangle  (Fig.  53),  the  effective  values  of 
the  three  q's  are  equal,  and  the  corresponding  instantaneous 
values  are  displaced  in  time  phase  by  120  degrees.  The  charg- 
ing current  per  unit  length  of  a  conductor  is  equal  to  the  rate 
of  change  of  the  corresponding  displacement  with  the  time,  or 

i  =  dq/dt (326) 

193 


194  THE  ELECTRIC  CIRCUIT  [AitT.  64 

But,  with  sinusoidal  voltages,  the  displacements  vary  also  ac- 
cording to  the  sine  law,  or 

5  =  Qrosm27r/f,     ......     (327) 

where  Qm  is  the  maximum  value  of  the  displacement  from  one  of 
the  conductors.     Substituting  this  value  of  q  in  eq.  (326),  we  find 

i  =  2  7rfQm  cos  2  TT/Y  ......     (328) 

Consequently,  the  amplitude  of  the  charging  current 

.......     (329) 


and  the  same  relation  holds  true  for  the  effective  values  of  the 
displacement  and  current.  It  is  to  be  noted  that  the  charging 
current  leads  the  flux  by  90  electrical  degrees.  Thus,  knowing 
the  displacement,  the  charging  current  can  be  calculated  from 
eq.  (329).  If  Qm  is  expressed  in  microcoulombs  per  kilometer, 
Im  is  in  microamperes  per  kilometer. 

The  actual  charging  current  which  flows  through  a  cross- 
section  of  the  conductor,  is  equal  to  that  necessary  to  supply  the 
displacement  between  this  cross-section  and  the  receiver  end  of 
the  line.  In  other  words,  the  charging  current  varies  along  the 
line,  from  a  maximum  at  the  generator  end  to  zero  at  the  receiver 
end.  If  the  effective  voltage  along  the  line  were  constant  in 
phase  and  magnitude,  the  amplitude  of  the  charging  current 
would  vary  according  to  a  straight-line  law.  In  reality,  the  volt- 
age varies  along  the  line,  due  to  its  resistance  and  inductance,  so 
that  the  variations  in  phase  and  amplitude  of  the  charging  cur- 
rent along  the  line  follow  a  much  more  complicated  law. 

The  influence  of  the  permittance  of  the  line  upon  its  voltage 
regulation  is  treated  in  Arts.  68  and  69  below.  The  problem  here 
is  a  preliminary  one;  namely,  with  a  given  size  and  arrangement 
of  conductors  in  a  three-phase  line,  to  find  the  permittance  per 
kilometer  of  the  equivalent  single-phase  line,  for  which  the  volt- 
age regulation  is  usually  calculated.  The  problem  is  solved  by 
applying  again  the  principle  of  superposition.  Each  conductor 
is  considered  as  forming  a  condenser  with  a  concentric  cylinder 
of  infinite  radius,  the  three  phases  being  star-connected  and  the 
three  cylinders  grounded.  The  vectors  of  the  star  and  delta  volt- 
ages are  shown  in  Fig.  53,  the  subscripts  1,  2,  3  referring  again 
to  the  conductors  A,  B,  and  C  respectively. 


CHAP.  XVII]      ELASTANCE  OF  THREE-PHASE  LINES 


195 


Applying  eq.  (290)  for  the  voltage  between  the  conductors  A 
and  B,  we  have,  for  instantaneous  values, 


en  =  ((79i/2  7r)'Ln  (6/a)  + 


TT)  Ln  (a/6), 


(330) 


where,  as  before,  the  spacing  is  denoted  by  6,  and  the  radii  of  the 
conductors  by  a.  The  first  term  on  the  right-hand  side  of  this 
equation  represents  the  action  of  system  A,  the  second  term  that 
of  system  B.  The  action  of  the  system  C  is  equal  to  zero,  be- 
cause, on  applying  eq.  (290)  for  this  system,  it  is  observed  that 
r  =  r',  on  account  of  the  symmetrical  spacing.  In  other  words, 


FIG.  53.     Electric  displacements  in  a  three-phase  line  with  symmetrical 
voltages  and  symmetrical  spacing. 


for  system  C  the  conductors  A  and  B  lie  on  the  same  equipoten- 
tial  cylindrical  surface.  The  preceding  equation  is  simplified  to 
ev,  =  (91  —  #>)  Sr,  where  S'  is  the  elastance  expressed  by  eq. 
(296),  that  is,  the  elastance  between  one  of  the  conductors  and 
the  plane  of  symmetry  00r,  as  if  the  third  conductor  did  not 
exist.  Owing  to  symmetry,  the  other  two  equations  are  similar; 
thus  we  have 

to -fa -,•) 

e23=  (92-93)5';        (331) 

631  =  (q*-qi)S'. 


196  THE  ELECTRIC  CIRCUIT  [AKT.64 

This  result  is  interpreted  graphically  by  Fig.  53,  remember- 
ing that  relations  which  hold  true  algebraically  for  instantane- 
ous values  of  sinusoidal  quantities,  hold  true  geometrically  for 
the  corresponding  vectors  of  these  quantities.  According  to  eqs. 
(331),  the  instantaneous  values  of  (qi  -  g2),  (92  -  Qs),  and  (q3  -  §1) 
are  in  phase  with  the  corresponding  voltages  ei2,  e23,  and  e3\.  For 
this  reason,  the  vectors  (Qi  -  Q2),  (Qz  -  Qa),  and  (Q3  -  Qi)  are 
drawn  in  phase  with  the  vectors  En,  E23,  and  E3i.  In  regard  to  the 
quantities  Qi,  Q2,  and  Q3,  we  know  that,  for  reasons  of  symmetry, 
they  are  equal  numerically  and  are  displaced  in  phase  relatively 
to  each  other  by  120  degrees.  Therefore,  they  must  be  repre- 
sented by  vectors  from  the  center  0  to  the  vertices  of  the  triangle 
MNP.  The  condition  is  then  fulfilled  that  each  side  of  this  tri- 
angle is  equal  to  the  difference  of  two  vectors  from  the  point  0. 

We  see  now  that  the  three  electric  displacements  Qi,  Q2,  and 
Q3  are  in  phase  with  the  corresponding  star-  or  Y-voltages  of  the 
system;  also,  from  the  similarity  of  the  triangles,  we  have  Eu/Ei 
=  (Qi  —  QaVQi,  with  corresponding  relations  for  the  other  two 
phases.  Consequently,  eqs.  (331)  are  reduced  simply  to 

El  =  Q£'; 

E2  =  QzS'; (332) 

E3  =  Q3S'. 

We  thus  arrive  at  the  following  important  conclusion:  The  dis- 
placement (and  consequently  the  charging  current)  per  phase  of  a 
three-phase  line  with  symmetrical  spacing  and  symmetrical  voltages 
is  equal  to  that  in  a  single-phase  line  with  the  same  conductors  and 
the  same  spacing,  provided  that  the  star  voltage  of  the  three-phase 
line  is  equal  to  that  between  one  conductor  and  the  neutral  plane  00' 
in  the  single-phase  line. 

As  explained  in  Art.  36,  an  equivalent  single-phase  line  is 
obtained  by  taking  one  conductor  of  the  three-phase  line  and 
assuming  the  transmission  voltage  to  be  equal  to  the  star  voltage 
of  the  actual  transmission  line;  the  return  conductor  is  supposed 
to  be  devoid  of  both  resistance  and  inductance.  The  preceding 
rule  gives  a  simple  method  for  finding  the  permittance  of  the 
equivalent  line;  namely,  the  permittance  of  the  equivalent  single- 
phase  line  is  equal  to  that  between  one  of  the  conductors  of  the  actual 
line  and  the  plane  of  symmetry  between  it  and  one  of  the  other  con- 
ductors, as  if  the  third  conductor  did  not  exist. 


CHAP.  XVII]     ELASTANCE  OF  THREE-PHASE  LINES  197 

The  calculation  of  the  charging  current  with  an  unsymmetrical 
spacing  of  conductors  is  much  more  involved,  and  is  explained  in 
the  next  article.  Fortunately,  however,  the  spacing  between  the 
conductors  affects  the  value  of  the  charging  current  but  little, 
with  the  usual  ratios  between  size  of  conductor  and  spacing. 
The  student  can  easily  verify  this  fact  by  consulting  any  available 
table  of  capacities  or  charging  currents  of  transmission  lines. 
The  reason  for  this  is  that  the  principal  part  of  the  elastance  be- 
tween two  small  conductors  occurs  near  the  conductors,  where 
the  flux  density  is  comparatively  high.  Consequently,  it  is  pos- 
sible in  practice  to  estimate  the  permittance  per  phase  of  a  three- 
phase  line  with  unsymmetrical  spacing,  by  finding  the  limits  of 
the  permittance  with  symmetrical  spacings.  For  instance,  let 
two  conductors  be  placed  on  a  cross-arm  and  the  third  on  top  of 
the  pole,  forming  an  isosceles  triangle.  Let  the  spacings  be 
2  m.  and  1.6  m.  respectively.  The  charging  currents  are  differ- 
ent in  the  three  conductors,  but  the  average  value  is  larger  than 
with  a  symmetrical  spacing  of  2m.,  and  smaller  than  with  a  sym- 
metrical spacing  of  1.6  m.  Having  found  the  charging  currents 
or  the  equivalent  permittances  for  these  two  spacings,  one  can 
assume  an  intermediate  value  by  interpolation,  or  else  take  one  of 
the  two  limits,  whichever  gives  the  more  unfavorable  operating 
conditions  of  the  line. 

It  is  rather  a  tedious  problem  to  estimate  the  influence  of  the 
ground  upon  the  charging  currents  in  a  three-phase  line.  The 
theory  is  simple,  the  ground  being  replaced  by  the  images  of  the 
three  conductors,  as  in  Fig.  51;  but  the  formulae  are  long  and  in- 
volved, because  the  effects  of  six  separate  systems  must  be  super- 
imposed. See  problem  3  in  the  next  article. 

Prob.  1.  Show  that  when  one  of  the  conductors  in  a  three-phase  line 
fails,  the  charging  current  in  the  other  two  conductors  drops  to  86.6  per 
cent  of  its  former  value.  Solution:  Let  C  be  the  permittance  between 
one  of  the  conductors  and  the  plane  of  symmetry  between  it  and  one  of 
the  other  conductors.  Then  the  charging  current  with  the  three  phases 
alive  is  kC(E/V^3),  where  E  is  the  line  voltage,  and  k  is  a  coefficient 
of  proportionality  with  which  we  are  not  concerned  here.  Operating 
single-phas^,  the  charging  current  is  kC(%E).  The  ratio  of  the  two  is 
0.5/  (I/  V3)  =  0.866. 

Prob.  2.  A  three-phase,  140-kv.,  25-cycle  transmission  line  consists  of 
conductors  2  cm.  in  diameter;  the  spacing  is  symmetrical  and  equal  to 
3.5  m.;  the  length  of  the  line  is  250  km.  What  is  the  total  reactive 


198  THE  ELECTRIC  CIRCUIT  [ART.  64 

power  necessary  to  keep  the  line  alive,  and  what  are  the  voltage  and  the 
permittance  per  kilometer  of  the  equivalent  single-phase  line? 

Ans.     7270  kva.;  80.8  kv.;  0.00947  mf.  per  km. 

Prob.  3.  A  three-phase  transmission  line  consists  of  conductors  18  mm. 
in  diameter,  suspended  all  three  in  the  same  vertical  plane,  at  a  distance 
of  2.4  m.  between  the  adjacent  conductors.  What  are  the  limits  of  the 
clastance  of  the  equivalent  single-phase  line? 

Ans.     100  and  113  megadarafs  per  km. 

Note:  The  proximity  of  the  two  limits  shows  that  it  is  sufficient  for 
practical  purposes  to  consider  the  symmetrical  spacing  only,  as  far  as 
the  dielectric  and  magnetic  effects  are  concerned.  Mr.  J.  G.  Pertsch,  Jr., 
has  called  the  author's  attention  to  the  fact  that,  with  certain  simplifying 
assumptions,  and  when  the  three  wires  are  transposed,  the  equivalent 
spacing  for  inductance  and  capacity  is  equal  to  the  geometric  mean  of 
the  three  actual  spacings,  or 

beq  =  ^6,2623631. 

In  the  case  under  consideration  the  equivalent  spacing  is  3.02  m.,  and  the 
corresponding  elastance  equals  105  mgd.  per  km. 

Prob.  4.  Extend  the  treatment  given  in  this  article  to  the  case  in 
which  the  three  delta  voltages  are  different  (Fig.  54),  and  show  that  the 

point  0  coincides  with  the  center 
of  gravity  of  the  triangle,  a  sym- 
metrical spacing  of  the  conductors 
being  presupposed  as  before.  Solu- 
tion: Equations  (331)  hold  true  as 
before,  and  the  sides  of  triangle 
MNP  are  parallel  to  those  of  123, 
but  the  point  0  cannot  be  deter- 
mined in  this  case  from  the  sym- 
metry of  the  figure.  Any  point  0 
within  the  triangle  123  gives  a  set 
of  star  voltages  Ei,  E2,  and  E3, 
which  will  produce  the  given  set  of 
delta  voltages;  but  there  is  only 
FIG.  54.  Electric  displacements  in  a  one  point  0  from  which  the  rays 
three-phase  line  with  unsymmctrical  ^°  t-ne  vertices  of  triangle  MNP 
voltages  and  symmetrical  spacing.  satisfy  condition  (325).  Since  the 

displacements    in    the    equivalent 

single-phase  lines  must  be  proportional  to  the  voltages,  condition  (325) 
requires  that  the  geometric  sum  of  Ei,  Et,  and  E3  shall  equal  zero. 
The  parallelogram  021'3  gives  the  geometric  sum  of  Et  and  E3  equal 
to  Or.  If  0  is  the  correct  point,  01'  must  be  equal  and  opposite  to 
01,  and  Om  =  JOl.  Similarly,  the  condition  must  be  fulfilled  that 
On  =  \  02,  and  Op  =  \  03.  It  is  known  from  elementary  geometry  that 
the  three  bisectors  of  a  triangle  divide  each  other  in  the  ratio  of  2  to  1, 
and  that  the  point  of  their  intersection  is  the  center  of  gravity  of  the 
triangle.  Hence  the  point  0  is  the  center  of  gravity  of  the  triangle  123. 


CHAP.  XVII]     ELASTANCE  OF  THREE-PHASE  LINES  199 

From  eqs.  (331)  we  again  derive  eqs.  (332),  and  finally  arrive  at  the 
same  conclusion  as  that  printed  in  italics  after  these  equations.  The 
three  star  voltages  being  different,  one  from  another,  the  three  charging 
currents  are  also  different,  each  leading  the  corresponding  Q  by  90  degrees. 
The  permittance  and  the  voltage  of  the  equivalent  single-phase  line  are 
also  different  for  each  phase,  in  spite  of  the  symmetrical  spacing  of  the 
conductors. 

Prob.  5.  For  a  given  three-phase  line  with  symmetrical  spacing  and 
voltages,  draw  the  electrostatic  field  for  the  instant  when  one  of  the 
delta  voltages  is  equal  to  zero;  also  make  three  drawings  of  the  field  at 
the  ends  of  intervals  A,  A  and  Tsz  of  a  cycle  later.  Use  the  principle  of 
superposition  explained  in  Prob.  5,  Art.  62,  and  apply  it  to  the  three 
component  systems,  A,  B,  and  C,  keeping  in  mind  the  relative  magni- 
tudes of  the  instantaneous  displacements. 

65.  Three-phase  Line  with   Unsymmetrical  Spacing.1      As 

is  mentioned  in  the  preceding  article,  the  calculation  of  charging 
currents  in  a  three-phase  line  with  unsymmetrical  spacing  is 
much  more  involved  than  with  symmetrical  spacing,  and  is  not 
of  much  practical  importance  at  present.  An  outline  of  it  is 
given  here  in  order  to  fix  more  firmly  in  the  student's  mind  the 
general  principle  of  superposition,  and  the  method  by  which  the 
results  are  derived  in  the  preceding  article.  Moreover,  the  in- 
fluence of  the  dielectric  is  becoming  more  and  more  important,  as 
the  transmission  voltages  and  the  lengths  of  transmission  lines 
are  increased.  The  time  may  come  when  the  exposition  given  in 
this  article  will  be  of  assistance  in  the  solution  of  practical  prob- 
lems. 

Let  the  three  spacings  be  denoted  by  612,  623,  and  631  respec- 
tively. Equations  (325)  to  (329)  inclusive  hold  true  as  with  a 
symmetrical  spacing,  but  eq.  (330)  now  becomes 

en  =  (<rgi/27r)  Ln  (&i2/a)  +  (<r?2/27r)  Ln  (a/6^)  + 

(<Tg3/27r)Ln(&23/&3i),       •     •     •     •'.--.     -     (333) 

because  the  effect  of  the  system  C  is  not  equal  to  zero  in  this  case. 
Similar  equations  may  be  written  for  623  and  631,  but  only  two 
equations  are  independent;  the  third  is  obtained  by  combining 
the  two  others,  because  the  third  voltage  in  a  delta  combination 
is  determined  by  the  other  two  voltages.  The  third  independent 
equation  is  (325),  and  these  three  equations  determine  the  three 
unknown  q's. 

1  This  article  may  be  omitted,  if  so  desired. 


200  THE  ELECTRIC   CIRCUIT  [ART.  65 

The  following  solution  of  these  equations  gives  an  insight  into 
the  physical  relations,  and  leads  to  a  result  which  is  convenient 
in  numerical  work.  The  last  term  on  the  right-hand  side  of 
eq.  (333)  is  usually  much  smaller  than  the  other  two  terms,  so 
that  it  may  be  conveniently  represented  in  the  form  of  a  cor- 
rection to  the  other  two,  thus  preserving  the  general  form  of  eq. 
(330).  Substituting  the  value  of  q3  from  eq.  (325)  into  (333),  we 
obtain 

e12  =  fai/2ir)  Ln  (bcl/a)  -  (aq2/2w}  Ln  (6c2/a),    .     (334) 

where  the  quantities, 

bcl  =  &12&31/& 

bc,  =  &23&12/&31,  ......        (335) 


may  be  called  the  corrected  spacings.  The  factors  by  which  the 
displacements  q\  and  q2  are  multiplied  in  eq.  (334)  are  familiar, 
since  they  are  of  the  same  form  as  the  right-hand  member  of 
eq.  (296).  It  will  be  recalled  that  eq.  (296)  expresses  the  elas- 
tance  between  one  conductor  and  the  plane  of  symmetry  of  a 
single-phase  line.  The  above-mentioned  factors  therefore  rep- 
resent the  elastances  of  single-phase  lines  having  the  corrected 
spacings  6ci  and  6c2  respectively.  Denoting  the  reciprocals  of  these 
elastances,  or  the  corrected  permittances,  by  C  with  the  corre- 
sponding subscripts,  eq.  (334)  and  the  two  similar  equations  for 
the  other  phases  are  reduced  to  the  form 


623  =  ?2/C2  -  q3/C3; 
e3i  =  q3/C3  - 

On  the  other  hand,  for  any  star  point  0,  no  matter  where 
located,  we  have  the  following  relation  between  the  delta  and 
star  voltages: 

612  =  ei  —  e2: 


(337) 


We  again  select  the  neutral  point  in  such  a  manner  that  each 
star  voltage  is  in  phase  with  the  corresponding  q;  that  is,  in 
phase  quadrature  with  the  corresponding  charging  current.  Then 
the  given  three-phase  system  is  directly  resolved  into  three  inde- 
pendent single-phase  lines,  and  our  problem  is  solved.  If  the 


CHAP.  XVII]      ELASTANCE  OF  THREE-PHASE  LINES  201 

point  0  is  so  selected,  then  by  comparing  eqs.  (336)  and  (337)  we 
have 


<?2/C2;     .......     (338) 


Substituting  the  values  of  the  g's  from  these  equations  into  eq. 
(325)  gives 

Ciei  +  C2e2  +  C3e3  =  0  ......     (339) 

This  is  the  condition  which  the  point  0  must  satisfy  if  eqs.  (338) 
are  to  hold  true.  Eliminating  e2  and  e3  from  eq.  (339)  by  means 
of  the  first  and  the  last  of  the  eqs.  (337),  and  solving  for  ei,  we 
obtain 

ei  =  e12C2/C  -  e31C3/C,  .....     (340) 
where 

C  =  Ci  +  C2  +  C3  .......     (341) 

As  mentioned  above,  Ci,  C2,  and  C3  are  the  permittances  per  unit 
length  between  one  wire  and  the  plane  of  symmetry,  for  the  cor- 
rected spacings  defined  by  eqs.  (335). 

Since  relations  which  hold  true  algebraically  for  instantaneous 
values  also  hold  true  geometrically  for  the  vectors  of  the  same 
quantities,  eq.  (340)  suggests  a 
simple  method  for  finding  graph- 
ically the  position  of  the  neutral 
point  0  in  the  vector  diagram 
(Fig.  55).  To  locate  0,  plot  1  k 
—  Eu(Cz/C}  in  the  direction  op- 
posite to  Eu,  and  W  =  E13(CS/C) 
parallel  to  E&.  Or  else,  the  prob- 
lem may  be  solved  analytically, 
using  either  the  orthogonal  or 
the  trigonometric  form  of  com- 
plex quantities.  Having  deter- 
mined the  position  of  0,  the  FIG.  55.  Electric  displacements  in 
three  star  voltages  become  known,  a  three-phase  system  with  unsym- 

,   ,v         ji  j-         j-  metrical  voltages  and  unsymmetrical 

and  then  the  corresponding  dis- 

placements are  found  from  eqs. 

(338).  The  charging  currents  are  determined  by  eq.  (329),  and 
are  in  leading  quadrature  with  the  corresponding  star  voltages. 
The  given  system  is  thus  resolved  into  three  independent  equiva- 


202  THE   ELECTRIC  CIRCUIT  [ART.  65 

lent  single-phase  systems  with  the  voltages  E1}  E2,  and  E3,  and 
the  permittances  Ci,  C2,  and  C3,  per  unit  length. 

Instead  of  using  the  treatment  given  above,  one  could  find  the 
equivalent  conductance  and  susceptance  by  using  a  method  anal- 
ogous to  that  employed  in  Art.  63  of  the  Magnetic  Circuit. 

Prob.  1.   Determine  the  actual  equivalent  elastances  in  problem  3  of 
the  preceding  article,  and  compare  them  with  the  assumed  limits. 

Prob.  2.   Extend  the  treatment  given  above  to  the  case  in  which  the 
cross-sections  of  the  three  conductors  are  different,  one  from  another. 

Prob.  3.  Show  how  to  estimate  the  influence  of  the  ground  upon  the 
charging  currents  in  a  three-phase  line,  using  the  method  of  successive 
approximations.  Solution:  Replace  the  conducting  ground  by  the  three 
images  A',  B',  and  C'  of  the  actual  conductors,  as  in  Fig.  51.  This  gives 
six  electric  systems  with  cylinders  at  infinity.  Applying  the  principle  of 
superposition  to  the  voltages  between  the  conductors  A — B  and  B — C, 
we  get,  by  analogy  with  eq.  (333) : 
e,2  =  (aqi/2  ,)  Ln  (612/a)  +  (aq./2,)  Ln  (a/612)  +  (aqs/2w)  Ln  (&„/&„) 

-  (aqi/2,)  Ln  (BA'/AA')  -  (aq,/2*)  Ln  (BB'/AB1) 

^   -(aq3/2,)Ln  (BC'/AC');  ,    (342) 

623  =  (<rjj/2  w) Ln (&ss/cO ~f"  (0q3/2ir)  Ln  (o/623)  -(-  (<rqi /27r)Ln  (6 31/612) 

-  (<rg2/2  TT)  Ln  (CB'/BB')  -  (ffq3/2  *)  Ln  (CC'/BC') 

From  these  two  equations,  together  with  eq.  (325),  the  three  unknown 
q's  can  be  evaluated.  By  a  method  similar  to  that  used  in  the  text 
above,  eqs.  (342)  are  conveniently  reduced  to  the  form 

«„  +  (a/2,)  fo,Ln  (BA'/AA')  +  ?2Ln  (BB'/AB')  +  q3Ln  (BC'/AC')}  ] 

=  ql/Cl-q,/Ct;  L  ,   , 

<?23  +  («/2,) [?2Ln (CB'/BB')  +  q3Lu (CC'/BC')+qiLn(CA'/BA')]  [  ^ 
=  qi/Cz  —  q^/Cz]  } 

where  the  three  C's  and  the  corrected  spacings  are  the  same  as  before, 
without  the  ground.  The  last  three  terms  on  the  left-hand  side  of  eqs. 
(343)  are  small  as  compared  to  612  and  e23,  and  represent  the  effect  of  the 
ground.  Therefore,  the  simplest  way  of  solving  these  equations  is  to 
neglect  the  correction  terms  in  the  first  approximation,  and  to  solve  for 
the  three  q's  exactly  as  explained  in  the  text  above,  by  finding  the  proper 
point  0  (Fig.  55).  The  correction  terms  may  be  said  to  modify  the 
values  of  en  and  e23  in  eqs.  (343).  Having  the  values  of  the  q's  in  the 
first  approximation,  the  corrections  are  calculated,  and,  being  added  to 
d2  and  e-i3,  give  new  values  of  the  latter,  say  e'n  and  e'23.  Having  thus 
modified  the  triangle  123  in  Fig.  55,  a  new  point  0  is  found,  and  new 
values  of  the  q's.  The  corrections  for  en  and  e23  can  now  be  determined 
more  accurately,  and  then  new  values  of  the  q's  found,  which  will  be  more 
nearly  correct  than  the  foregoing  ones.  In  this  way,  the  influence  of  the 
ground  can  be  estimated  with  any  desired  degree  of  accuracy,  without 
solving  long  and  involved  simultaneous  equations. 


CHAPTER  XVIII 

DIELECTRIC  REACTANCE  AND  SUSCEPTANCE  IN 
ALTERNATING-CURRENT  CIRCUITS 

66.  Dielectric  Reactance  and  Susceptance.  Let  a  condenser 
of  permittance  C,  or  elastance  S,  be  connected  across  an  alternat- 
ing-current line  of  voltage  E  and  frequency  /.  Let  any  instanta- 
neous value  of  the  voltage  be  denoted  by  e,  where  e  =  Em  sin  2  vft; 
then  the  corresponding  instantaneous  displacement  in  the  dielec- 
tric is 

q  =  eC  =  e/S (344) 

This  displacement  varies  according  to  the  sine  law  and  is  in  phase 
with  the  voltage,  because  q  is  at  every  instant  proportional  to  e. 
The  charging  current  flowing  from  the  line  into  the  condenser  is 
at  any  instant  equal  to  the  rate  of  change  of  q  with  the  time,  or 

i  =  dq/dt  =  2  wfCEm  cos  2wft (345) 

It  will  be  seen  from  this  equation  that  the  charging  current  leads 
the  voltage  by  90  degrees,  as  has  already  been  explained  in  Art.  48 
above.  The  amplitude  of  the  charging  current  is 

Im  =  2irfCEm (346) 

It  may  thus  be  said  that  a  permittance  C  connected  across  a 
source  of  voltage,  of  frequency  /,  is  equivalent  to  a  susceptance 

b=-Im/Em=-2TfC (347) 

The  minus  sign  is  necessary  because  the  current  is  leading,  while 
with  a  magnetic  susceptance  it  is  lagging.  In  other  words,  by 
using  the  minus  sign  in  the  case  of  dielectric  susceptance,  and  the 
plus  sign  for  magnetic  susceptance,  it  is  possible  to  extend  the 
formulae  deduced  in  Chapters  8  and  9  to  alternating-current  cir- 
cuits containing  dielectrics. 

In  the  preceding  formulae  C  is  in  farads,  S  in  darafs,  q  in 
coulombs,  and  6  in  mhos.  If  C  is  expressed  in  microfarads  and 
S  in  megadarafs,  eq.  (347)  becomes 

b  =  -27T/C  X  10-«  =  -2  TT/  X  10-VS  mhos.  .         (348) 


204  THE  ELECTRIC  CIRCUIT  [ART.  66 

The  corresponding  dielectric  reactance  is 

x  =  - 106/(2  7T/C)  =  - 106  S/(2  TT/)  ohms.      .     .     (349) 

The  dielectric  susceptance  is  equal  to  -  2  irfC  only  when  there  is 
no  resistance  in  series  with  the  condenser.  When  a  condenser  is 
connected  in  series  or  in  parallel  with  an  ohmic  resistance,  the 
treatment  is  analogous  to  that  of  a  magnetic  inductance  in  combi- 
nation with  a  resistance;  that  is,  equivalent  series  and  parallel  com- 
binations are  used,  as  explained  in  Art.  27.  To  give  a  detailed 
treatment  here  would  simply  be  to  repeat  what  has  already  been 
explained  in  the  above-mentioned  article.  The  only  difference  is 
that  expressions  (348)  and  (349)  are  used  in  place  of  (107)  and 
(86),  and  the  currents  are  leading,  while  with  magnetic  reactance 
they  are  lagging  with  respect  to  the  impressed  voltage. 

In  some  circuits  both  magnetic  and  dielectric  susceptances 
are  connected  in  parallel.  They  are  simply  added,  taking  into  con- 
sideration their  opposite  signs.  For  instance,  a  magnetic  suscept- 
ance of  7  mhos  in  parallel  with  a  dielectric  susceptance  of  5  mhos 
is  equivalent  to  a  net  magnetic  susceptance  of  2  mhos.  A  simi- 
lar rule  is  applied  when  magnetic  and  dielectric  reactances  are  con- 
nected in  series. 

With  these  explanations,  the  student  will  have  no  difficulty 
in  dealing  with  any  combination  of  resistances,  condensers,  and 
inductance  coils  in  an  alternating-current  circuit. 

Prob.  1.  A  condenser  of  7.3  mf.  permittance  is  connected  across  a 
500-volt,  60-cycle  supply.  What  are  the  susceptance  and  the  charging 
current? 

Ans.  b  =  -0.002754  mho;  I  =j  1.377  amp.,  the  voltage  being  the 
reference  vector. 

Prob.  2.  The  condenser  in  the  preceding  problem  is  shunted  by  a 
non-inductive  resistance  of  750  ohms.  Find  the  total  current  and  the 
power-f actor.  Solution:  The  current  through  the  resistance  is  =  500/750 
=  0.6667  amp.;  tan  <f>  =  1.377/0.6667  =  2.065;  cos  <t>  =  43.58  per  cent 
(leading).  Total  current  =  0.6667/0.4358  =  1.53  amp. 

Prob.  3.  The  condenser  and  the  resistance  in  the  preceding  problem 
are  connected  in  series,  instead  of  in  parallel.  What  is  the  equivalent 
parallel  combination? 

Ans.     Cp  =  1.387  mf.;    rp  =  926  ohms. 

Prob.  4.  The  voltage  at  the  receiver  end  of  a  25-cycle,  single-phase 
transmission  line  is  45  +  j'57  kv.;  the  load  current  is  178  +  j  69  amp. 
The  series  magnetic  impedance  of  the  line  is  32  +  j  68  ohms,  and  its 
capacity  is  4.24  mf.  Calculate  the  generator  current  and  voltage.  For 
purposes  of  calculation,  one  half  of  this  capacity  can  be  assumed  to  be 


CHAP.  XVIII]  DIELECTRIC  REACTANCE  205 

connected  across  the  generator  end  of  the  line,  the  other  half  across  the 
receiver  end.  Solution:  The  dielectric  susceptance  at  the  receiver  end 
of  the  line  is  -2  *  X  25  X  2.12  X  KT6  =  -0.333  X  10~3  mho.  The 
corresponding  charging  current  is 

j  0.333  X  10~3  (45000  + .;  57000)  =  - 19  +  j  15  amp. 

Consequently  the  total  line  current  is  159  +  j  84  amp.  The  line  drop  is 
(159  +  j  84)  (32  +  j  68)  =  -624  +  j  13500  volts.  The  generator  volt- 
age is  44.38  +  j  70.5  kv.  The  charging  current  at  the  generator  end  is 
j  0.333  (44.38  +  j  70.5)  =  -23.5  +  j  14.79  amp.  The  generator  current 
is  135.5  +  j  98.8  amp. 

Prob.  5.  Explain  the  physical  reason  why  a  dielectric  susceptance  in- 
creases with  the  frequency,  while  a  magnetic  susceptance  is  inversely 
proportional  to  it. 

Prob.  6.  Investigate  the  influence  of  a  condenser  in  a  circuit  to  which 
a  non-sinusoidal  voltage  is  applied;  give  a  treatment  similar  to  that  in 
Art.  23.  Show  that  the  presence  of  an  elastance  accentuates  higher 
harmonics  in  the  current,  while  an  inductance  tends  to  diminish  them. 
Make  the  physical  reason  for  this  difference  clear  to  yourself. 

67.  Current  and  Voltage  Resonance.  Let  a  condenser  be 
connected  in  parallel  with  a  pure  reactance  coil,  across  an  alter- 
nating-current line.  Let  the  current  through  the  condenser  be 
5  amp.,  leading,  and  that  through  the  coil,  3  amp.,  lagging. 
Then  the  total  current  supplied  from  the  generator  is  2  amp., 
leading.  Thus,  we  have  the  paradox  that  the  resultant  current  is 
smaller  than  either  of  its  components.  It  is  even  possible  to 
adjust  the  permittance  and  inductance  to  such  values  as  to  make 
the  leading  and  lagging  components  equal,  in  which  case  the  gen- 
erator current  is  zero.  This  condition  is  called  current  resonance. 
When  the  line  current  is  reduced,  to  zero,  total  or  perfect  reson- 
ance takes  place;  otherwise  the  resonance  is  called  partial.  The 
condition  for  perfect  resonance  is  that  the  lagging  current  shall  be 
equal  to  the  leading  current,  or,  what  is  the  same,  the  dielectric 
susceptance  must  be  numerically  equal  to  the  magnetic  suscept- 
ance. Thus,  if  there  is  no  resistance  in  either  circuit, 

24G-1/04L), 

from  which 

2irfVCL  =  l (350) 

From  this  equation,  any  one  of  the  three  quantities  /,  (7,  and  L 
can  be  determined,  when  the  other  two  are  given.  Condition 
(350)  may  be  fulfilled  for  the  frequency  of  one  of  the  higher  bar- 


206  THE  ELECTRIC   CIRCUIT  [AKT.  67 

monies  of  an  e.m.f.  wave,  in  which  case  we  have  partial  resonance 
for  the  fundamental  wave,  and  perfect  resonance  for  one  of  the 
harmonics.  If  such  is  the  case,  the  line  current  does  not  contain 
this  harmonic,  although  it  may  be  present  to  a  considerable 
amount  in  the  two  branch  currents. 

From  the  point  of  view  of  energy,  current  resonance  consists 
in  a  periodic  transformation  of  the  potential  energy  of  the  elec- 
trostatic field  into  the  kinetic  energy  of  the  magnetic  field,  and 
vice  versa.  When  the  current  is  at  its  maximum,  the  energy  of 
the  magnetic  field  of  the  reactance  coil  is  also  a  maximum.  But 
at  this  moment  the  voltage,  and  consequently  the  electrostatic 
displacement,  are  equal  to  zero,  so  that  the  whole  energy  of  the 
circuit  is  in  the  magnetic  field.  One  quarter  of  a  cycle  later,  the 
displacement  and  the  stored  energy  in  the  condenser  are  at  a 
maximum,  but  the  current  and  the  magnetic  field  are  equal  to 
zero.  At  intermediate  moments,  the  energy  is  contained  partly 
in  the  electrostatic,  and  partly  in  the  magnetic  field.  When 
condition  (350)  is  satisfied,  the  maxima  of  the  two  energies  are 
numerically  equal,  and  the  system  "  oscillates  "  freely  in  the 
electrical  sense,  in  a  manner  analogous  to  the  swinging  of  a  pendu- 
lum. The  generator  merely  maintains  the  necessary  frequency, 
and  supplies  the  izr  loss.  Without  this  loss,  it  would  not  be 
necessary  to  have  the  generator  at  all;  the  oscillations,  once 
started,  would  continue  indefinitely  at  the  proper  frequency. 
When  the  two  energies  are  not  equal,  there  must  be  a  cyclic  ex- 
change of  energy  between  the  generator  and  one  of  the  branches; 
namely,  the  one  whose  storage  capacity  for  energy,  at  the  gener- 
ator frequency,  is  larger  than  that  of  the  other  branch.  We  then 
have  partial  current  resonance. 

The  presence  of  resistance  in  either  branch  obscures  the  effect 
of  resonance  to  some  extent,  leaving,  however,  its  general  char- 
acter unchanged.  The  best  way  to  see  the  influence  of  resistance 
is  to  replace  each  impedance  by  its  equivalent  parallel  combina- 
tion. We  then  have  two  pure  susceptances  with  reactive  currents, 
and  two  conductances,  the  currents  through  which  are  in  phase 
with  the  line  voltage.  The  energy  supplied  to  the  conductances 
is  converted  into  heat,  and  thus  does  not  enter  into  the  electrical 
oscillations. 

Let  now  a  dielectric  reactance  be  connected  in  series  with  a 
magnetic  reactance,  across  an  alternating-current  line.  The  cur- 


CHAP.  XVIII]  DIELECTRIC  REACTANCE  207 

rent  through  the  two  devices  is  the  same,  and  may  be  taken  as 
the  reference  vector.  Let  the  dielectric  reactance  be  such  as  to 
produce  across  the  condenser  a  drop  of  1000  volts,  lagging  behind 
the  current  by  90  electrical  degrees.  Let  the  voltage  across  the 
reactance  coil  be  equal  to  900  volts,  leading  the  current  by  90 
degrees.  With  these  conditions,  the  total  line  voltage  is  equal  to 
100  volts,  lagging  behind  the  current  by  90  degrees.  Thus,  with 
a  line  voltage  of  only  100,  it  is  possible  to  produce  partial  voltages 
of  1000  and  900  respectively.  This  condition  is  called  voltage 
resonance.  When  the  two  reactances  in  series  are  equal,  we  have 
complete  voltage  resonance;  otherwise  the  resonance  is  partial. 
The  student  will  readily  see  that  the  condition  for  complete  volt- 
age resonance  is  also  expressed  by  eq.  (350).  In  this  case,  the 
presence  of  resistance  has  no  effect  upon  the  correctness  of  the 
equation.  By  reading  again  the  foregoing  discussion  of  current 
resonance,  and  applying  it  to  voltage  resonance,  the  points  of 
similarity  and  the  differences  between  the  two  will  be  easily 


One  has  to  be  on  guard  against  possible  resonance  and  a 
dangerous  rise  in  potential  in  the  operation  of  transmission  lines 
and  extended  cable  systems,  because  there  the  presence  of  per- 
mittance and  inductance  offers  favorable  conditions  for  surges 
between  the  dielectric  and  magnetic  energies.  These  surges  either 
produce  large  currents  which  open  the  circuit-protecting  devices 
and  interrupt  the  service,  or  the  potential  is  raised  to  a  value  at 
which  the  insulation  of  the  system  is  broken  down.  With  a  clear 
understanding  of  the  principle  of  interchange  of  energy  explained 
above,  the  student  ought  to  be  able  to  follow  without  difficulty 
special  works  on  the  subject.1 

Prob.  1.  A  magnetic  reactance  of  65  ohms  is  connected  in  parallel 
with  a  permittance  of  73.6  mf.,  across  a  2200-volt,  25-cycle  circuit.  De- 
termine the  total  current,  and  the  component  currents,  through  the  react- 
ance and  through  the  condenser. 

Ans.  33.85  —  25.44  =  8.41  amp.  (lagging).  This  is  a  case  of  partial 
current  resonance,  the  total  current  being  smaller  than  one  of  its  com- 
ponents. 

1  See  W.  S.  Franklin,  Electric  Waves;  C.  P.  Steinmetz,  Electric  Dis- 
charges, Waves,  and  Impulses;  also  his  larger  work  on  Transient  Electric 
Phenomena  and  Oscillations.  Some  elementary  experiments  and  curves  of 
resonance  will  be  found  in  V.  Karapetoff's  Experimental  Electrical  Engineer- 
ing, Vol.  II,  Arts.  440  to  445. 


208  THE  ELECTRIC  CIRCUIT  [ART.  68 

Prob.  2.  The  pcrmittor  and  the  reactance  coil  given  in  the  preceding 
problem  are  connected  across  the  same  line  in  series,  instead  of  in  parallel. 
Find  the  total  current  and  the  component  voltages. 

AIMS.  102.3  amp.  (leading) ;  2200  =  8850  -  6650  volts.  This  is  a 
case  of  partial  voltage  resonance,  the  voltage  drop  across  each  of  the 
two  devices  being  larger  than  the  applied  voltage. 

Prob.  3.  The  elastance  of  a  60-cycle  underground  cable  system  is 
equal  to  11.5  kilodarafs;  at  what  value  of  the  inductance  in  the  circuit  is 
resonance  of  the  seventh  harmonic  to  be  feared? 

Ans.     1.65  millihenry. 

68.  Voltage  Regulation  of  a  Transmission  Line,  Taking  Its 
Distributed  Permittance  into  Account.  The  voltage  regulation 
of  a  transmission  line,  disregarding  its  permittance,  is  treated  in 
Art.  33.  The  value  of  the  permittance  of  a  single-phase  line  is 
deduced  in  Art.  60,  while  in  Chapter  17  the  effect  of  the  charging 
current  in  a  three-phase  line  is  considered,  and  it  is  shown  how 
to  calculate  the  permittance  of  the  equivalent  single-phase  line. 
The  inductance  of  transmission  lines  is  treated  in  Chapter  11  of 
the  author's  Magnetic  Circuit.  It  remains  now  to  show  how  to 
determine,  for  a  given  load,  the  relation  between  the  generator 
and  receiver  voltages  of  an  equivalent  single-phase  line,  knowing 
its  constants;  viz.,  the  values  of  the  distributed  resistance,  mag- 
netic reactance,  and  dielectric  susceptance. 

Let  the  total  resistance  of  the  equivalent  single-phase  line  be 
r  ohms,  and  its  magnetic  reactance  x  ohms.  Then  the  series 
impedance  of  the  line  is 

Z  =  r  +  jx (351) 

Let  the  dielectric  susceptance  of  the  line  be  b  mhos,  where  &,  ac- 
cording to  eq.  (347),  is  a  negative  quantity;  and  let  the  leakage 
conductance  to  the  ground  be  g  mhos.  Then  the  shunted  ad- 
mittance of  the  line  is 

Y  =  g-jb (352) 

The  leakage  conductance  is  due  to  imperfect  insulation  of  the 
line,  and  may  also  be  made  to  take  into  account  the  corona  loss, 
if  any  exists.  The  value  of  g  can  only  be  estimated,  and  in  most 
cases  may  be  safely  neglected.  It  is  introduced  here  in  order  to 
obtain  a  more  general  result,  at  the  same  time  making  the  ex- 
pressions for  Z  and  Y  symmetrical. 

Since  Y  is  uniformly  distributed  along  the  line,  the  current 
changes  as  the  distance  from  the  generator  increases;  and  there- 


CHAP.  XVIII]  DIELECTRIC  REACTANCE  209 

fore  it  is  necessary  to  consider  the  electrical  relations  in  an  in- 
finitesimal length  ds,  at  some  intermediate  point  of  the  line.  Let 
the  line  voltage  at  this  point  be  E,  and  the  line  current,  /.  The 
series  impedance  of  the  element  ds  is  (Z/l)  ds,  and  its  shunted 
admittance  is  (Y/l)  ds,  where  I  is  the  total  length  of  the  line. 
Let  dE  be  the  increment  in  the  voltage  in  the  length  ds,  and  let 
dl  be  the  corresponding  increment  in  the  line  current  due  to  the 
shunted  admittance.  We  have  then 

dE  =-I(Z/l)ds,      .....     (353) 
and 

dl  =  -  E(Y/l)  ds  ......     (354) 

The  minus  sign  is  needed  on  the  right-hand  side  of  eq.  (353), 
because  the  drop  in  voltage  /  (Z/l)  ds  causes  a  decrement  in  E. 
Likewise  the  charging  current  E(Y/l)ds  causes  a  decrement  in 
the  line  current. 

Equations  (353)  and  (354)  contain  two  dependent  variables, 
E  and  /.  To  eliminate  /,  we  divide  both  sides  of  eq.  (353)  by  ds 
and  take  the  derivative  with  respect  to  s.  The  result  is 


Substituting  the  value  of  dl/ds  from  eq.  (354),  we  obtain 

d*E/ds2  =  EZY/l*  .......     (355) 

This  is  a  differential  equation  of  the  second  order  for  E.  We  shall 
omit  the  solution  of  it  and  give  only  the  result,  for  two  reasons: 
first,  because  most  students  are  not  familiar  with  the  methods  of 
integration  of  differential  equations;  and  secondly,  because  the 
solution  is  most  conveniently  expressed  in  hyperbolic  functions  of 
a  complex  variable,  a  form  of  function  unknown  to  most  students 
of  engineering.1  Fortunately,  even  for  the  longest  transmission 

1  The  simple  theory  of  hyperbolic  functions  and  the  solution  of  eq.  (355) 
may  be  found,  among  others,  in  the  following  works  and  articles  :  McMahon, 
Hyperbolic  Functions;  Dr.  Kennelly,  Applications  of  Hyperbolic  Functions  to 
Electrical  Engineering;  Dr.  Steinmetz,  Transient  Electric  Phenomena;  Fender 
and  Thomson,  "  The  Mechanical  and  Electrical  Characteristics  of  Transmission 
Lines,"  Trans.  Amer.  Inst.  Electr.  Engrs.,  Vol.  30  (1911);  W.  E.  Miller,  "  Hyper- 
bolic Functions  and  Their  Application  to  Transmission  Line  Problems,"  General 
Electric  Review,  Vol.  13  (1910),  p.  177;  M.  W.  Franklin,  "Transmission  Line 
Calculations,"  ibid.,  p.  74.  For  a  proof  of  expansion  (356),  see  Blondel  and 
Le  Roy,  "Calcul  des  Lignes  de  Transport  d'Energie  a  Courants  Alternatifs  en 
tenant  compte  de  la  Capacity  et  de  la  Perditance  Reparties,"  La  Lumiere 


210  THE  ELECTRIC   CIRCUIT  [ART.  68 

lines  built  or  projected,  the  solution  can  be  represented  with  suffi- 
cient accuracy  by  a  few  terms  of  an  infinite  series,  as  follows: 
E,  =  E2(l  +  %YZ  +  &  F2Z2  +  etc.) 

+  7,Z(1  +  |  YZ  +  TW  F2Z2  +  etc.).      (356) 

In  this  equation  E\  is  the  generator  voltage,  E2  the  receiver  volt- 
age, the  same  as  in  Art.  33,  and  /2  is  the  load  current.  Both  F 
and  Z  are  known  complex  quantities,  and  therefore  their  product 
and  the  square  of  the  product  .are  also  known.  The  terms  in- 
volving F2Z2  are  negligibly  small  in  many  cases. 

Instead  of  eliminating  /  from  eqs.  (353)  and  (354),  E  may  be 
eliminated  by  a  similar  process,  giving  a  differential  equation  for 
/,  analogous  to  eq.  (355).  The  solution  of  this  equation  is 

!i  =  /«(!  +  I  YZ  +  ^  F2Z2  +  etc.) 

+  E2Y  (1  +  |  YZ  +  Tio  F2Z2  +  etc.),      (357) 

where  /i  is  the  generator  current,  and  1%  the  load  current. 

The  general  form  of  eqs.  (356)  and  (357)  is  the  same  as  that  of 
the  corresponding  equations  in  Art.  33,  and  in  Chapters  11,  12,  and 
13,  so  that  the  methods  of  calculation  indicated  there  are  appli- 
cable here,  with  self-evident  modifications. 

Neglecting  the  leakage  conductance  g  in  eq.  (352),  and  using 
the  value  of  permittance  given  in  Art.  60  above,  also  the  value  of 
inductance  from  Art.  61  of  the  Magnetic  Circuit,  we  find  that 

YZ  =  (Z/1000)2  (-v+jw),     ....    (358) 
and  consequently 

F2Z2  =  (Z/1000)4  (*>2  -  w*  -  2jvw),  .     .     .     (359) 
where 

v  =  0.09514  (0.1  /)2  [0.46  +  0.05/log  (b/d)],      .     (360) 
and 

w  =  0.1515  /r'/log(6/a) (361) 

In  these  expressions,  I  is  the  length  of  the  line  in  kilometers,  and 
r'  is  the  resistance  per  kilometer  of  one  conductor,  in  ohms.  With 

Elcclrique,  Vol.  7,  2nd  Scries  (1909),  p.  355;  also  J.  F.  H.  Douglas,  "Trans- 
mission Line  Calculations,"  Electrical  World,  Vol.  55  (1910),  p.  1066;  arid  Dr. 
Stoinmetz,  Engineering  Mathematics,  p.  204.  The  best  tables  of  hyperbolic 
functions  are  those  published  by  the  Smithsonian  Institution;  briefer  tables 
will  be  found  in  McMahon's  book  and  in  the  General  Electric  Review,  Vol.  13, 
supplement  to  No.  5.  See  also  Seaver's  Mathematical  Handbook,  pp.  85  and 
266. 


CHAP.  XVIII]  DIELECTRIC  REACTANCE  211 

the  extreme  values  of  the  ratio  b/a  of  say  10  and  1000,  the  value 
of  log  (b/a)  varies  within  the  narrow  limits  of  1  to  3,  so  that  the 
second  term  in  the  brackets  in  eq.  (360)  is  comparatively  small. 
In  practice,  the  value  of  the  whole  expression  in  the  brackets  in 
formula  (360)  is  usually  between  0.48  and  0.50.  This  fact  is 
taken  advantage  of  in  numerical  calculations  which  do  not  require 
particular  accuracy. 

Prob.  1.   Check  the  numerical  coefficients  in  formulae  (360)  and  (361). 

Prob.  2.  For  a  given  receiver  voltage,  calculate  the  generator  voltage, 
at  no  load  and  at  full  load,  for  some  very  long  transmission  line,  the 
dimensions  of  which  are  taken  from  a  descriptive  article. 

Prob.  3.  Solve  problem  2  by  the  use  of  tables  of  hyperbolic  functions, 
following  the  method  indicated  in  one  of  the  references  in  the  footnote. 
Compare  the  results  with  those  obtained  in  problem  2,  and  make  clear  to 
yourself  the  relative  simplicity,  and  the  limits  of  accuracy,  of  the  series 
when  one,  two,  or  three  terms  are  used. 

69.  Approximate  Formulae  for  the  Voltage  Regulation  of  a 
Transmission  Line,  Considering  Its  Permittance  Concentrated  at 
One  or  More  Points.  Instead  of  treating  the  permittance  of  a 
transmission  line  in  the  correct  manner  described  in  the  preceding 
article,  it  is  sometimes  assumed  to  be  concentrated  at  one  or  more 
points  along  the  line.  The  calculation  of  voltage  regulation  then 
becomes  similar  to  the  treatment  in  Chapters  9  to  13.  There  is 
no  particular  advantage  in  this  approximate  treatment  as  far  as 
the  simplicity  of  numerical  computations  is  concerned,  because 
the  formulse  obtained  are  similar  to  eq.  (356).  It  is  advisable, 
however,  for  the  student  to  deduce  such  formulae  in  order  to  see 
for  himself  that  the  form  of  eq.  (356)  is  a  rational  one;  more- 
over, this  gives  him  one  more  exercise  in  the  use  of  complex 
quantities. 

(a)  The  simplest  assumption  is  to  consider  one  half  of  the 
line  permittance  (and  leakage,  if  any)  concentrated  at  the  gener- 
ator end  of  the  line,  the  other  half  at  the  receiver  end.  The  load 
current  is  in  this  case  apparently  increased  by  the  current  E2  •  %  Y 
through  the  permittance  \  Y  connected  in  parallel  with  the  load, 
so  that  the  total  receiver  current  is  equal  to  /2  +  \  E2Y.  Hence, 
the  generator  voltage  is 


#1  =  E2(l  +  i  FZ)  +  72Z.  .     (362) 


212  THE  ELECTRIC   CIRCUIT  [ART.  69 

Comparing  this  formula  with  eq.  (356),  we  see  that  the  principal 
terms  are  identical,  the  difference  being  in  the  additional  terms 
containing  higher  powers  of  YZ.  If  the  influence  of  the  line  per- 
mittance is  small,  for  instance  in  short  lines,  the  results  calculated 
by  means  of  both  formulae  differ  from  each  other  but  very  little. 
There  is  no  reason,  however,  why  the  accurate  expansion  (356) 
should  not  be  used  in  all  cases,  taking  as  many  terms  as  are  re- 
quired in  a  given  problem. 

The  generator  current,  with  the  capacity  concentrated  at  both 
ends,  is 

7,  =  7,  +  i  YE,  +  i  YElt 

or,  substituting  the  value  of  E\  from  eq.  (362), 

!i  =  J»U  +  *YZ)  +  E2Y(1  +  J  YZ).     .     .     (363) 

This  formula  is  similar  to  eq.  (357),  and  differs  from  it  only  in 
the  values  of  the  coefficients  of  the  minor  terms. 

(b)  The  line  permittance  and  leakage  may  also  be  concen- 
trated at  the  middle  point  of  the  line,  in  which  case  a  diagram  of 
connections  is  obtained  similar  to  Fig.  42,  except  that  the  suscept- 
ance  is  dielectric  and  not  magnetic.     Introducing  the  voltage  at 
the  center  of  the  line  as  an  auxiliary  unknown  quantity,  and 
eliminating  it  from  the  result,  we  obtain 

Ei  =  E2(l  +  i  YZ)  +  /aZ(l  +  I  YZ),    .     .     (364) 
and 

Ii  =  |2(1  +  £  YZ)  +  E,Y (365) 

(c)  A  closer  approximation  is  obtained  by  assuming  a  part  of 
the  line  permittance  concentrated  at  the  middle  point,  and  the 
rest  at  both  ends  of  the  line.     The  fractions  of  the  total  permit- 
tance to  be  assigned  to  these  three  points  are  determined  from 
Simpson's  Rule  for  approximate  integration;  namely,  according 
to  this  "parabolic  "  rule, 

yave  =  [l/(3  n)]  [yQ  +  4  (yi  +  ys  +  etc.  +  yn^) 

+  2  (yt  +  y,  +  etc.  +  yn-2)  +  yn], (366) 

where  yave  is  the  average  ordinate  of  a  given  curve,  n  is  the  num- 
ber of  equal  parts  into  which  the  total  width  of  the  curve  is  sub- 
divided, and  t/o,  yi,  etc.,  are  the  actual  ordinates  at  the  points  of 
division.  In  the  above  formula,  n  must  be  an  even  number. 
Let  the  given  curve  represent  some  arbitrary  distribution  of  the 


CHAP.  XVIII]  DIELECTRIC  REACTANCE  213 

permittance  along  the  line,  and  let  n  =  2.     The  foregoing  formula 
gives 

<72'),     ....     (367) 


where  the  C's  are  marked  with  the  prime  sign  to  indicate  that 
they  refer  to  unit  length  of  the  line.  But  in  reality  the  permit- 
tance is  uniformly  distributed  over  the  length  of  the  line,  so  that 
Ca,e'  =  Co'  =  Ci  =  C2f.  Multiplying  both  sides  of  eq.  (367)  by 
the  length  I  of  the  line,  we  obtain 

C  =  iC+fC  +  |C  ......     (368) 

This  means  that  two  thirds  of  the  total  permittance  must  be  con- 
centrated at  the  middle  of  the  line,  and  one  sixth  at  each  end.1 

With  this  distribution  of  permittance  it  is  again  convenient 
to  introduce  the  voltage  at  the  center  of  the  line  as  an  auxiliary 
quantity.  The  relation  between  the  load  voltage  and  the  gener- 
ator voltage  is  calculated  in  the  well-known  manner,  by  adding 
the  voltage  drop  in  the  line  to  the  load  voltage.  The  result  is 

Ei  =  E.2(l  +  ±YZ  +  &  Y*Z2)  +  I2Z(1  -H  FZ);  .     (369) 
!i  =  /i(l  +  *  YZ  +  A  F2Z2)  +  E2Y(l  +  &  YZ 

+  *}-irr2Z2)  ............     (370) 

These  formulae  come  closer  to  eqs.  (356)  and  (357)  than  those 
obtained  in  the  preceding  two  approximations. 

Prob.  1.  Check  formulae  (364)  and  (365)  by  actually  performing  the 
algebraic  transformations. 

Prob.  2.  Check  formula  (369)  and  (370)  by  actually  performing  the 
algebraic  transformations. 

Prob.  3.  If  it  be  desired  to  have  the  permittance  concentrated  at 
five  equidistant  points  along  the  line,  show  that  according  to  Simpson's 
Rule  one  sixth  of  the  total  permittance  must  be  placed  in  the  middle, 
one  twelfth  at  each  end,  and  the  rest  at  one  quarter  and  three  quarters 
of  the  length  of  the  line. 

1  This  result  has  been  first  indicated  by  Dr.  Steinmetz,  in  his  Alternating- 
Current  Phenomena,  in  the  chapter  on  "  Distributed  Capacity." 


APPENDIX. 


THE   AMPERE-OHM    SYSTEM    OF   UNITS. 

THE  ampere  and  the  ohm  can  be  now  considered  as  two 
arbitrary  fundamental  units  established  by  an  international  agree- 
ment. Their  values  can  be  reproduced  to  a  fraction  of  a  per 
cent  according  to  detailed  specifications  adopted  by  practically 
all  civilized  nations.  These  two  units,  together  with  the  centi- 
meter and  the  second,  permit  the  determination  of  the  values  of 
all  other  electric  and  magnetic  quantities.  The  units  of  mass  and 
of  temperature  do  not  enter  explicitly  into  the  formulae,  but  are 
contained  in  the  legal  definition  of  the  ampere  and  the  ohm. 
The  dimension  of  resistance  can  be  expressed  through  those  of 
power  and  current,  according  to  the  equation  P  =  PR,  but  it  is 
more  convenient  to  consider  the  dimension  of  R  as  fundamental, 
in  order  to  avoid  the  explicit  use  of  the  dimension  of  mass  [M]. 
Besides,  there  is  no  direct  proof  that  the  physical  dimensions  of 
electric  power  are  the  same  as  those  of  mechanical  power.  All 
we  know  is  that  the  two  kinds  of  power  are  equivalent  one  to 
the  other. 

For  the  engineer  there  is  no  need  of  using  the  electrostatic  or 
the  electromagnetic  units;  for  him  there  is  but  one  ampere-ohm 
system,  which  is  neither  electrostatic  nor  electromagnetic.  The 
ampere  has  not  only  a  magnitude,  but  a  physical  dimension  as 
well,  —  a  dimension  which,  with  our  present  knowledge,  is  fun- 
damental; that  is,  it  cannot  be  reduced  to  a  combination  of  the 
dimensions  of  length,  time,  and  mass  (or  energy).  Let  the  dimen- 
sion of  current  be  denoted  by  [I]  and  that  of  resistance  by  [R]; 
let  the  dimensions  of  length  and  time  be  denoted  respectively  by 
the  commonly  recognized  symbols  [L]  and  [T].  The  magnitudes 
and  dimensions  of  all  other  electric  units  can  be  expressed  through 
these  foui,  as  shown  in  the  following  table.  For  the  correspond- 
ing expressions  of  the  magnetic  units  in  the  ampere-ohm  system, 
see  Appendix  I  to  the  author's  Magnetic  Circuit. 

215 


216 


THE  ELECTRIC  CIRCUIT 


TABLE  OF  ELECTRIC  UNITS,   AND  THEIR  DIMENSIONS  IN 
THE  AMPERE-OHM   SYSTEM 


Symbol  and  Formula 

Quantity 

Dimensions 

Name  of  the  Unit 

j 

Current 

[I] 

Ampere 

U=I/A 

Current  density 

[IL-2J 

Ampere  per  square 

centimeter 

Q=IT 

Quantity  of  electricity 
and  dielectric  flux 

[IT]* 

Coulomb     (ampere- 
second) 

D  =  Q/A 

Dielectric  flux  density 

[ITL-2] 

Coulomb  per  square 

centimeter 

E=IR 

Voltage,   difference   of 

[IR] 

Volt 

potential,  or  e.m.f. 

G=E/l 

Voltage  gradient,  elec- 

[IRL-i] 

Volt  per  centimeter 

tric  intensity,  or  di- 

electric stress 

T 

Resistance 

j 

X  =  2irfL 

Reactance 

[R] 

Ohm 

2=(r2+x2)i 

Impedance 

) 

g=l/rp 

Conductance 

j 

b  =  l/xp 

Susceptance 

[    [R-1] 

Mho 

y=  1/2=  (02-|-£>2)* 

Admittance 

) 

P  =  G/U 

Resistivity 

[RL] 

Ohm  per  centimeter 

cube 

y=U/G=l/P 

Conductivity 

[R-^-1] 

Mho  per  centimeter 

cube 

S=E/Q 

Elastance 

[RT-1] 

Daraf 

C  =  Q/E=l/S 

Permittance  (capacity) 

[R-iT] 

Farad 

a  =  G/D 

Elastivity 

[RT-iL] 

Daraf  per  centime- 

ter cube 

K=D/G=l/a 

Permittivity 

[R-iTL-1] 

Farad  per  centime- 

ter cube 

P=EI 

Power 

[PR] 

Watt 

P'  =  P/V 

Density  of  power 

[PRL-3] 

Watt  per  cubic  cen- 

timeter 

W'-\GD 

Stored  electric  energy 
Density  of  electric  en- 

[PRT] 
[PRTL-3] 

Joule  (watt-second) 
Joule  per  cubic  cen- 

ergy 

timeter 

F=W/l 

Force 

[PRTL-1] 

Joulecen 

L"WI" 

Inductance 

[RT] 

Henry 

*  These  are  also  the  dimensions  of  the  electric  pole  strength.  The  con- 
cept of  pole  strength  is  of  no  use  in  electrical  engineering,  and,  in  the  author's 
opinion,  its  usefulness  in  physics  is  more  than  doubtful.  The  whole  elemen- 
tary theory  of  electrostatics  can  and  ought  to  be  built  up  on  the  idea  of 
stresses  and  displacements  in  the  dielectric,  as  is  done  in  this  work. 


THE  AMPERE-OHM  SYSTEM  217 

Other  units  of  more  convenient  magnitude  are  easily  created 
by  multiplying  the  tabulated  units  by  powers  of  10,  or  by  adding 
the  prefixes  milli-,  micro-,  kilo-,  mega-,  etc. 

A  study  of  the  physical  dimensions  of  the  electric  and 
magnetic  quantities  is  interesting  in  itself,  and  gives  a  better 
insight  into  the  nature  of  these  quantities.  Moreover,  for- 
mute  can  be  checked  and  errors  detected  by  comparing  phys- 
ical dimensions  on  both  sides  of  the  equation.  Let,  for  instance, 
a  formula  for  energy  be  given, 

W  =  aQDl/K, 

where  a  is  a  numerical  coefficient.  Substituting  the  physical  di- 
mensions of  all  the  quantities  on  the  right-hand  side  of  the  equa- 
tion from  the  table  below,  the  result  will  be  found  to  be  of  the 
dimensions  of  energy.  This  fact  adds  to  one's  assurance  that 
the  given  formula  is  theoretically  correct. 

A  slight  irregularity  in  the  system  as  outlined  above  is  caused 
by  the  use  of  the  kilogram  as  the  unit  force,  because  it  leads  to 
two  units  for  energy  and  torque,  viz.,  the  kilogram-meter  and 
the  joule;  1  kg.-meter  =  9.806  joules.  Force  ought  to  be  measured 
in  joules  per  centimeter  length,  to  avoid  the  odd  multiplier.  Such 
a  unit  is  equal  to  about  10.2  kg.,  and  could  be  properly  called 
the  joulecen  (=  107  dynes).  There  is  not  much  prospect  in  sight 
of  introducing  this  unit  of  force  into  practice,  because  the  kilo- 
gram is  too  well  established  in  common  use.  The  next  best 
thing  to  do  is  to  derive  formulae  and  perform  calculations,  when- 
ever convenient,  in  joulecens,  and  to  convert  the  results  into 
kilograms  by  multiplying  them  by  g  =  9.806. 

Thus,  leaving  aside  all  historical  precedents  and  justifica- 
tions, the  whole  system  of  electric  and  magnetic  units  is  re- 
duced to  this  simple  scheme:  In  addition  to  the  centimeter,  the 
gram,  the  second,  and  the  degree  Centigrade,  two  other  funda- 
mental units  are  recognized,  the  ohm  and  the  ampere.  All 
other  electric  and  magnetic  units  have  dimensions  and  values 
which  are  connected  with  those  of  the  fundamental  six  in  a 
simple  and  almost  self-evident  manner,  as  shown  in  the  table 
above. 

To  appreciate  fully  the  advantages  of  the  practical  ampere- 
ohm  system  over  the  C.G.S.  electrostatic  and  electromagnetic 
systems,  one  has  only  to  compare  the  dimensions,  for  instance,  of 


218 


THE  ELECTRIC  CIRCUIT 


current  density  and  voltage  gradient  in  these  three  systems,  as 
shown  below. 


The  Am- 
pere-ohm 
System 

C.G.S.  Electro- 
magnetic System 

C.G.S.  Electro- 
static System 

Dimension  of  current  density.  .  . 
Dimension  of  voltage  gradient.  . 

IL-2 
IRL"1 

L-»M»T-V* 
L»M*T-V 

L-»M*T-'«* 

L-*M*T-\-* 

BIBLIOGRAPHY. 


ALTERNATING-CURRENT  PHENOMENA,  by  Chas.  P.  Steinmetz. 
THEORETICAL  ELEMENTS  OF  ELECTRICAL  ENGINEERING,  by  Chas.  P.  Stein- 
metz. 

ALTERNATING  CURRENTS,  by  Bedell  and  Crehore. 
VECTORS  AND  VECTOR  DIAGRAMS,  by  Cramp  and  Smith. 
REVOLVING  VECTORS,  by  G.  W.  Patterson. 
ELECTRICAL  ENGINEERING,  by  Thomalen. 

DlE   WlSSENSCHAFTLICHEN   GRUNDLAGEN  DER  ELEKTROTECHNIK, 

by  Benischke. 

DIE  WECHSELSTROMTECHNIK,  by  E.  Arnold. 
PROBLEMS  IN  ELECTRICAL  ENGINEERING,  by  Waldo  V.  Lyon. 
ELECTRICAL  PROBLEMS,  by  Hooper  and  Wells. 

THE  ELEMENTS  OF  ELECTRICAL  ENGINEERING,  by  Franklin  and  Eaty. 
PRINCIPLES  OF  ELECTRICAL  ENGINEERING,  by  H.  Pender. 
MODERN  VIEWS  OF  ELECTRICITY,  by  Oliver  Lodge. 
ELECTRIC  WAVES,  by  W.  S.  Franklin. 
ELEMENTS  OF  ELECTROMAGNETIC  THEORY,  by  S.  J.  Barnett. 
KAPAZITAT  UNO  INDUKTIVITAT,  by  Ernst  Orlich. 


219 


INDEX 

PAGE 

Addition  and  subtraction  of  sinusoidal  currents  and  voltages 40 

Admittance,  definition  of 76 

expressed  as  a  complex  quantity  or  operator 89 

Admittances  in  parallel 76 

in  series 80 

Air,  dielectric  strength  of 165 

permittivity  of 151 

Alternating-current  power  when  current  and  voltage  are  in  phase 45 

currents,  advantages  of 31 

currents,  polyphase 99 

Ampere-ohm  system  of  units 3,  215 

Amplitude  factor,  definition  of 51 

Analogy,  hydraulic,  to  an  inductive  circuit 65 

hydraulic,  to  Ohm's  law 2 

hydraulic,  to  the  dielectric  circuit 145 

hydraulic,  to  the  flow  of  electricity 25 

mechanical,  to  a  charged  dielectric 159 

thermal,  to  Ohm's  law 2 

thermal,  to  the  flow  of  electricity 24 

Apparent  power,  definition  of 56 

Average  value  of  alternating  current  or  voltage 50 

Cable,  elastance  of  single-core 171 

insulation  resistance  of 26 

Cables,  grading  insulation  of 174 

Capacity,  see  Elastance  and  Permittance. 

electrostatic,  definition  of 147 

specific  inductive 151 

Charging  current  of  condenser 203 

current  of  transmission  line,  see  also  Elastance 193 

currents  of  three-phase  line  with  symmetrical  spacing 196 

currents  of  three-phase  line  with  unsymmetrical  spacing 201 

Circle  coefficient  of  induction  motor 138 

diagram  of  induction  motor  or  transformer 136 

Circuit,  alternating-current 31 

dielectric 143 

dielectric,  hydraulic  analogue 145 

direct-current 1 

polyphase 99 

221 


222  INDEX 


Coefficient,  leakage,  of  induction  motor 138 

of  self-induction  see  Inductance. 

temperature  of  electric  resistivity 5 

Complex  expression  for  admittance 

expression  for  impedance 

quantity,  definition  of 85 

Component,  energy,  of  current  or  voltage 56 

reactive,  of  current  or  voltage 56 

Condenser,  charging  current  of 203 

definition  of 143 

Conductance  and  resistance,  how  related,  in  an  A.C.  circuit 79 

definition  of 2 

dielectric 109 

Conductances,  addition  of 8 

Conductivity,  definition  of 14 

Conductor,  definition  of 1 

of  variable  cross-section 22 

unit,  definition  of 13 

Continuous  current,  see  Current,  direct. 

Core  loss  of  transformer Ill 

Corona,  electrostatic 167 

Current,  alternating 31 

density,  definition  of 15 

direct 1 

due  to  non-sinusoidal  voltage 71 

effective  value  of  alternating 48 

energy  and  reactive  components  of 56 

primary,  of  induction  motor 126 

radial  flow  of 26 

refraction,  law  of 28 

resonance 205 

transient,  in  opening  and  closing  a  circuit 71 

Currents,  polyphase  alternating 99 

Cycle  of  alternating  wave,  definition  of 33 

Cylinders,  elastance  between  two  large  parallel 188 

Daraf,  definition  of 148 

Delta-connected  three-phase  system 105 

Dielectric  circuit 143 

conductance 169 

elastivity  of 152 

energy  stored  in 158 

flux  density 154 

flux,  refraction  of 164 

hysteresis 169 

nature  of 144 

permittance  of 147 


INDEX  223 

PAGE 

Dielectric,  permittivity  of 151 

reactance 204 

strength 164 

stress 156 

susceptance 203 

Dimensions  of  units,  table  of 216 

Dispersion  factor  of  induction  motor 138 

Displacement,  electric,  illustrated 144 

Disruptive  voltage,  see  Dielectric  strength. 

Effective  value  in  terms  of  harmonics 54 

value  of  variable  current 49 

values  of  alternating  currents  and  voltages,  definition  of 48 

Elastance  between  concentric  spheres 175 

between  small  spheres 179 

between  two  large  parallel  cylinders 188 

definition  of 148 

of  a  single-core  cable 171 

of  a  single-phase  line 176 

of  a  three-phase  line  with  symmetrical  spacing 196 

Elastances,  addition  of 149 

Elastivity,  definition  of 152 

Electric  displacement  illustrated 144 

intensity,  definition  of 16 

intensity  in  the  dielectric  circuit 155 

power 10 

Electromotive  force,  see  Voltage. 
Electrostatic,  see  also  Dielectric. 

capacity,  definition  of 147 

corona 167 

field,  nature  of 143 

Energy  component  of  current  or  voltage 56 

converted  into  heat 10 

density  of 158 

stored  in  dielectric 158 

stored  magnetic 62 

unit  of  electrical 11 

Equipotential  surfaces  denned 22 

Equivalent  resistance,  definition  of 7 

series  and  parallel  circuits 78 

sine-wave,  definition  of 53 

Exciting  admittance  of  transformer Ill 

Exponential  expressions  for  vectors  and  operators 97 

Farad,  definition  of 147 

Field,  electrostatic,  see  Electrostatic  field. 

Fleming's  method  for  calculating  the  effective  value  of  an  irregular  curve.  52 


224  INDEX 

PAGE 

Flux  density,  dielectric   154 

dielectric,  see  Dielectric  flux. 

Form  factor,  definition  of 51 

Fourier  series 43 

Frequency  of  alternating  current  or  voltage,  definition  of 33 

Gradient,  voltage,  in  the  dielectric  circuit 155 

voltage,  definition  of 16 

Ground,  influence  upon  the  charging  currents  in  a  three-phase  line  ....  202 

influence  upon  the  elastance  of  a  single-phase  line 180 

Harmonics,  definition  of 41 

effects  of  elastance  and  inductance  on 205 

Heaviside,  Oliver,  nomenclature  of 152 

Henry,  definition  of 62 

Heyland  diagram  of  induction  motor  or  transformer 136 

Homopolar  machine 31 

Horse-power,  English,  defined 10 

metric,  defined 10 

Hydraulic  analogue  of  inductive  circuit 65 

analogue  of  the  dielectric  circuit 145 

Hysteresis,  dielectric : . . .  169 

Images,  Kelvin's  method  of  electric 180 

Impedance,  definition  of 67 

equivalent,  of  transformer 116 

expressed  as  a  complex  quantity  or  operator 88 

Impedances  in  parallel 80 

in  series 68 

Inductance,  definition  of 60 

influence  of,  with  non-sinusoidal  voltage 69 

Induct  ion  motor,  approximate  analytical  treatment 125 

characteristics  with  locked  rotor 123 

circle  coefficient  or  dispersion  factor 138 

circle  or  Heyland  diagram  of 136 

equivalence  to  a  polyphase  transformer 122 

equivalent  electrical  diagram  of 122 

exact  analytical  treatment 139 

input  per  phase 127 

magnetomotive  forces  in 124 

maximum  output  of 131 

primary  current  and  power-factor 126 

pull-out  torque  of 130 

secondary  resistance  and  reactance  reduced  to  primary.  133 

slip,  calculation  of 126 

slip,  defined 123 

squirrel-cage  rotor 134 


INDEX  225 

PAGE 

Induction  motor,  starting  torque  of 129 

torque  of 127 

Inductive  reactance,  see  Reactance. 

Inertia  as  an  analogue  to  inductance 60 

Insulation,  see  also  Dielectric. 

condenser  type 175 

grading  of 174 

Intensity,  electric,  definition  of 16 

electric 155 

factor,  illustrated 10 

Irregular  paths,  resistance  and  conductance  of 27 

Joule,  definition  of 11 

relation  of,  to  thermal  units 11 

Joulecen,  definition  of 217 

Joule's  law 10 

Kelvin's  law  of  economy 15 

Kelvin's  method  of  electric  images 180 

Kirchhoff's  laws 17 

Law,  Joule's 10 

Kirchhoff's  first 18 

Kirchhoff's  second 19 

of  current  refraction 28 

of  flux  refraction 163 

of  economy,  Kelvin's 15 

of  minimum  resistance 27,  160 

Ohm's,  synopsis  of 1 

Leakage  conductance  of  transmission  line 208 

factor  of  the  induction  motor 138 

Lehmann,  Dr.,  method  of  finding  resistance  of  irregular  conductor 28 

Lehmann,  Dr.,  method  of  mapping  irregular  field 162 

Line,  see  Transmission  line. 

Magnetizing  current  of  transformer Ill 

Mean,  see  Average. 

Mesh  connection  of  polyphase  system 101 

Mho,  definition  of 2 

Minimum  resistance,  law  of 160 

Motor,  induction,  see  Induction  motor. 

Neutral  points  of  polyphase  system 104 

Nomenclature xii 

Notation xiii 

Ohm's  law,  for  an  infinitesimal  conductor 25 

hydraulic  analogy  to 2 


226  INDEX 

PAGE 

Ohm's  law,  synopsis  of 1 

thermal  analogy  to 2 

Operator,  admittance 89 

impedance 88 

Operators  expressed  as  exponential  functions 97 

polar  expressions  for 93 

Output,  maximum,  of  induction  motor 131 

Parallel  connection  of  conductors 7 

connection  of  impedances 80 

connection  of  susceptances  and  conductances 77 

Performance  characteristics  of  the  induction  motor 122 

characteristics  of  the  transformer 108 

characteristics  of  the  transmission  line 94,  208 

Permittance,  see  also  Elastance. 

definition  of 147 

distributed,  of  transmission  lines 208 

Permittances,  addition  of 149 

Permittivity,  definition  of 151 

relative «. 151 

Phase  angle,  definition  of 34 

Phase  displacement  expressed  by  projections  of  vectors 91 

Polar  coordinates,  vectors  and  operators  in 93 

Polyphase  system,  definition  of 99 

system,  neutral  points  of 104 

Power,  alternating-current,  when  current  and  voltage  are  in  phase 45 

apparent 56 

as  double-frequency  function 59 

average,  of  non-sinusoidal  waves 57 

electric 10 

expressed  by  projections  of  vectors 91 

expression  for  average  value  in  alternating-current  circuit 56 

practical  unit  of 10 

real 56 

Power-factor,  definition  of 56 

of  induction  motor 126 

with  non-sinusoidal  waves 58 

Projections  of  vectors,  addition  and  subtraction  of 82 

Quadratic  mean  value 50 

Quantity  factor,  illustrated 10 

Quarter-phase  system,  star-  and  mesh-connected 101 

Radial  flow  of  current 26 

Rayleigh,  Lord,  method  of  finding  permittance  of  dielectrics  of  irregular 

shape  161 

method  of  finding  resistance  of  irregular  conductor 28 

Reactance  and  susceptance  in  an  A.C.  circuit 79 


INDEX  227 

PAGE 

Reactance,  definition  of 64 

dielectric 204 

equivalent,  of  transformer 118 

inductive 63 

leakage,  of  transformer 110 

secondary,  of  induction  motor  reduced  to  primary 133 

Reactive  component  of  current  or  voltage 56 

Refraction  of  current 28 

of  dielectric  flux .\ 164 

Regulation,  see  also  Voltage  regulation. 

speed,  of  induction  motor 123 

voltage,  of  the  transformer 108 

voltage,  of  the  transmission  line 94 

Resistance  and  conductance  in  A.C.  circuits 79 

and  temperature,  relation  between 5 

definition  of 1 

equivalent,  definition  of 7 

equivalent,  of  transformer 118 

law  of  minimum 160 

secondary,  of  induction  motor  reduced  to  primary 133 

Resistances,  addition  of 8 

Resistivity,  definition  of 13 

Resonance,  current 205 

voltage 207 

Series  connection  of  admittances 80 

connection  of  conductors 7 

connection  of  impedances 68 

Series-parallel  combination  of  permittances  and  elastances 150 

combination  of  resistances 9 

Sine-wave,  definition  of 32 

definition  of  equivalent 53 

of  current  or  voltage 31 

represented  by  a  vector 36 

Single-phase  line,  effect  of  the  ground  upon  the  elastance  of 180 

elastance  of 176 

equations  of  lines  of  force  and  equipotential  surfaces 184 

Sinusoidal  currents  and  voltages,  rule  for  addition  and  subtraction  of ...  40 

Slip,  calculation  of 

of  induction  motor  defined 123 

Specific  capacity,  see  Permittivity,  relative, 
resistance,  see  Resistivity. 

Spheres,  elastance  between  concentric 175 

elastance  between  small 179 

equations  of  lines  of  force  and  equipotential  surfaces  between .  .  187 

Square  root  of  mean  square  value,  defined 49 

Star  connection  of  polyphase  system 101 


228  INDEX 

PAGB 

Steinmetz,  Dr.  C.  P.,  symbolic  notation  of 83 

Stream  lines,  definition  of 22 

Superposition,  principle  of 177 

Susceptance  and  reactance,  how  related,  in  an  A.C.  circuit 79 

definition  of 75 

dielectric 203 

Susceptances  in  parallel 75 

Symbols,  list  of xiii 

System,  four-wire,  two-phase 99 

three-wire,  two-phase 100 

polyphase,  definition  of •. 99 

quarter-phase,  star-  and  mesh-connected 101 

three-phase,  delta-connected 105 

three-phase,  V-  and  T-connected 107 

three-phase,  Y-connected 103 

T-connected  three-phase  system 107 

Temperature  coefficient 5 

Thermal  resistance,  definition  of 2 

Three-phase  line,  influence  of  the  ground  upon  the  charging  currents. .  .  .  202 

with  symmetrical  spacing,  elastance  and  charging  current  of. ...  196 

with  unsymmetrical  spacing,  charging  currents  of 201 

Three-phase  system,  delta-connected 105 

V-  and  T-connected 107 

Y-connected 103 

Time  constant  of  electric  circuit 72 

Torque  of  induction  motor 127 

pull-out,  of  induction  motor 130 

starting,  of  induction  motor 129 

Transformer,  constant-potential,  definition  of 108 

core  loss  of Ill 

equivalent  impedance  of 116 

equivalent  resistance  and  reactance  of 118 

exciting  admittance  of Ill 

leakage  reactance  of 110 

magnetizing  current  of Ill 

ohmic  drop  in 110 

reactive  drop  in 110 

vector  diagram  of 113 

voltage  ratio  of 109 

voltage  regulation 108,  115, 120 

Transient  current  in  opening  and  closing  a  circuit 71 

Transmission  line,  see  also  Three-phase  line  and  Single-phase  line. 

leakage  conductance  of 208 

voltage  regulation  of 94 

voltage  regulation,  taking  account  of  distributed  per- 
mittance                                                 .  208 


INDEX  229 

PAGE 

Tube  of  current,  meaning  of 23 

Two-phase,  four-wire  system 99 

three-wire  system ; 100 

Unit  conductor,  definition  of 13 

of  conductance 2 

of  electrical  energy 11 

of  resistance 1 

Units,  C.  G.  S.  and  practical  systems 3 

international  electrical 3 

table  of  names  and  dimensions  of 216 

the  ampere-ohm  system 215 

V-connected,  three-phase  system 107 

Vector,  definition  of 36 

diagrams,  examples  of 100,  102,  113 

used  to  represent  a  sine-wave . 36 

Vectors,  addition  and  subtraction  of 37 

addition  and  subtraction  of  projections  of 82 

expressed  as  exponential  functions 97 

in  polar  coordinates 93 

Voltage,  effective  value  of  alternating 48 

energy  and  reactive  components  of 57 

•  gradient  in  the  dielectric  circuit 155 

gradient,  definition  of 16 

gradient,  rupturing  values  of 165 

regulation  of  the  transformer 108,  115,  120 

regulation  of  the  transmission  line 94 

regulation  of  transmission  line,  with  permittance  concentrated.  211 

regulation  of  transmission  line,  with  permittance  distributed.  . .  208 

resonance 207 

Watt,  definition  of 10 

Wave  form  of  alternating  current  or  voltage 51 

representation  of  irregular 41 

Y-connected,  three-phase  system 103 

Yrneh,  definition  of 148 


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